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ELEMENTARY   ALGEBRA 


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ELEMENTAKY  ALGEBRA 


USE   OF   SCHOOLS   AND  COLLEGES 


BY 

CHARLES   SMITH,  M.A. 

AUTHOR  OF  "a  treatise  ON  ALGEBRA,"  "AN  ELEMENTARY 
TREATISE  ON  CONIC  SECTIONS,"  ETC. 

f 

BE  VISED  AND  ADAPTED   TO  AMERICAN  SCHOOLS 
BY 


IRVING   STRINGHAM,  Ph.D. 

prof: 

\ 


professor  of  MATHEMATICS  AND  SOMETIME  DEAN 
IN   THE  UNIVERSITY  OF  CALIFORNIA 


^    THE 


THE   MACMILLAN   COMPANY 

LONDON:  MACMILLAN  &  CO.,  Ltd. 
1902 

AU  righta  reserved 


'\c>X 


COPYEIGHT,   1894, 

By  MACMILLAN  AND  CO. 

Copyright,  1900, 
By  the  MACMILLAN   COMPANY. 


Set  up  and  electrotyped  June,  1894.  Reprinted  September,  1894; 
February,  August,  1895;  January,  September,  1897;  January,  Au- 
gust, 1898;   October,  1899. 

New  revised  edition  August,  1900  ;  April,  1902. 


NorfajDOlf  ^regg 

J.  S.  Gushing  &  Co.  —  Berwick  &  Smith 

Norwood  Mass.  U.S.A. 


PREFACE  TO  THE  AMERICAN  EDITION. 


The  transition  from  the  traditional  algebra  of  many  of 
our  secondary  schools  to  the  reconstructed  algebra  of  the 
best  American  colleges  is  more  abrupt  than  is  necessary 
or  creditable.  This  lack  of  articulation  between  the 
work  of  the  schools  and  the  colleges  emphasizes  the 
need  of  a  fuller  and  more  thorough  course  in  elementary 
algebra  than  is  furnished  by  the  text-books  now  most 
commonly  used.  It  is  with  the  hope  of  supplying  this 
new  demand  that  an  American  edition  of  Charles  Smith's 
Elementary  Algebra  is  published ;  a  work  whose  excel- 
lencies, as  represented  in  former  editions,  have  been 
recognized  by  able  critics  on  both  sides  of  the  Atlantic. 

In  the  rearrangement  of  the  work  and  in  its  adaptation 
to  American  schools  many  changes  have  been  made,  too 
many  to  be  noted  in  a  short  preface,  and  a  considerable 
amount  of  new  subject-matter  has  been  introduced.  The 
following  are  innovations  of  some  importance : 

Chapter  I.,  consisting  of  a  series  of  introductory 
lessons,  is  wholly  new,  and  Chapter  XIII.  is  partly 
new  and  partly  transferred  from  Chapter  XXVIII.  of 
the  second  edition.  Horner's  synthetic  division  is  made 
prominent  in  the  chapter  on  division,  an  early  introduc- 
tion to  quadratic  equations  finds  its  appropriate  place  in 
the  chapter  on  factoring,  the  binomial  theorem  for  posi- 


'29;^ 


VI  •  PREFACE. 

tive  integral  exponents  is  demonstrated  by  elementary- 
methods  in  the  chapter  on  powers  and  roots,  and  the 
chapter  on  surds  has  been  enlarged  by  a  short  discussion 
of  complex  numbers.  Some  new  collections  of  examples 
have  been  introduced,  and  several  of  the  older  lists  have 
been  extended. 

As  thus  reconstructed,  the  book  constitutes  a  rounded 
course  in  what  may  be  called  the  newer  elementary  alge- 
bra, and  includes  the  subject-matter  specified  by  nearly 
all  American  colleges  as  the  requirement  for  admission. 
It  will  prove  especially  helpful  to  students  preparing  for 
such  colleges  as  are  using  Mr.  Smith's  Treatise  on  Algebra 
for  advanced  work. 

Those  who  are  familiar  with  the  second  edition  will 
notice  that  the  chapters  on  permutations  and  combina- 
tions, on  logarithms  and  exponentials,  on  the  binomial 
theorem  for  negative  and  fractional  indices,  on  scales  of 
notation,  and  on  cube  root  have  been  omitted.*  .  .  . 

I  am  indebted  to  my  colleagues  of  the  mathematical 
department  of  the  University  of  California  for  valuable 
suggestions,  but  especially  to  Professor  Haskell  and 
Dr.  Hengstler  for  contributions  to  subject-matter  and 
for  reading  many  of  the  proof-sheets. 

Special  thanks  are  due  to  Mr.  Smith,  for  allowing  free 

scope  for  this  revision. 

IRVING  STRINGHAM. 
University  of  California, 
May,  1894. 

*  Chapters  on  logarithms,  exponentials,  and  inequalities  have  been 
introduced  into  this  second  briefer  edition;  and  in  the  second  complete 
edition  new  chapters  on  the  theory  of  equations  are  now  incorporated. 

L  S.  — Paris,  December,  1899. 


PREFACE  TO  THE  SECOND  EDITION. 


In  response  to  many  appeals  from  teachers,  a  series 
of  short  chapters  introductory  to  the  theory  of  equations 
—  Chapters  XL.  to  XL VIII.  inchisive  —  are  added  to 
this  second  edition.  In  the  interest  of  a  better  classifica- 
tion of  subject-matter,  the  chapter  on  determinants  and 
the  chapter  on  scales  of  notation,  of  the  first  edition, 
have  been  interchanged  and  renumbered  in  the  second; 
but  with  this  exception  there  has  been  no  renumbering 
of  chapters  or  articles. 

Necessarily,  in  a  brief  survey  of  the  theory  of  equa- 
tions, many  important  topics  are  omitted,  and  the  sub- 
jects that  are  included  are  discussed  but  meagrely.  For 
a  more  thorough  and  systematic  reading  of  this  impor- 
tant theory,  especially  in  its  modern  phases,  the  student 
will  have  recourse  to  the  larger  treatises. 

Irving  Stringham. 
Paris,  September,  1899. 


Til 


CONTENTS. 

•o*— ^ — 

HAPTBB  PAGE 

I.     Introductory  Lessons 1 

II.     The  Language  of  Algebra.     Definitions 17 

Positive  and  Negative  Quantities 27 

III.  Addition 32 

Subtraction 37 

Brackets 43 

IV.  Multiplication 46 

Product  of  Multinomial  Expressions 56 

Continued  Products 67 

V.     Division 71 

Division  by  Multinomial  Expressions 75 

Homer's  Synthetic  Division 78 

Miscellaneous  Examples  I 88 

VI.     Simple  Equations 92 

VII.     Problems 101 

VIII.     Simultaneous  Equations  of  the  First  Degree 109 

Elimination  by  Addition  and  Subtraction 110 

Other  Methods  of  Elimination 114 

Equations  with  Three  Unknown  Quantities 119 

IX.     Problems 122 

Miscellaneous  Examples  II 128 

X.     Factors 132 

Factors  of  ax"^  -^  bx  +  c 142 

Factors  by  Rearrangement  of  Terms 147 

Quadratic  Equations 151 

iz 


CONTENTS. 


CHAPTER 

XI. 

XII. 

XIII. 


XIV. 


XV. 


XVI. 


XVII. 

XVIII. 

XIX. 


XX. 


PAGE 

Highest  Common  Factors 156 

Lowest  Common  Multiples 168 

Miscellaneous  Theorems  and  Examples 174- 

Mathematical  Induction. . .  < 174 

Factor  Theorem.    Remainder  Theorem 179 

Fractions 186 

Reduction  to  a  Common  Denominator 193 

Addition  of  Fractions 194 

Multiplication  of  Fractions 201 

Division  of  Fractions 202 

Reciprocal.     Infinity 203 

Two  Important  Theorems 208 

Equations  with  Fractions 216 

Simultaneous  Equations 221 

Miscellaneous  Examples  III 229 

Quadratic  Equations 234 

General  Properties 239 

Irrational  Equations 245 

Relations  between  the  CoeflBcients  and  the  Roots 249 

Special  Forms 251 

Equations  with  Given  Roots 253 

Equations  of  Higher  Degree  than  the  Second. . ; 260 

Simultaneous  Equations  of  the  Second  Degree 266 

Problems 279 

Miscellaneous  Examples  IV ....  285 

Miscellaneous  Equations 289 

Involution 297 

The  Binomial  Theorem 302 

Evolution 309 

Square  Root 310 


CONTENTS. 


XI 


CHAPTER 

XXI. 
XXII. 


XXIII. 


^XIV. 

XXV. 

XXVI. 


XXVII. 
XXVIII. 

XXIX. 


XXX. 
XXXI. 
XXXII. 


XXXIII. 
XXXIV. 

XXXV. 
XXXVI. 


PAGE 

Fractional  and  Negative  Indices 319 

^urds ...;...  328 

Complex  Quantities 337' 

Ratio 344 

Proportion 349 

Variation 357 

Miscellaneous  Examples  V 363 

Arithmetical  Progression 367 

Geometrical  Progression 380 

Harmonical  Progression 393 

Other  Simple  Series 398 

Miscellaneous  Examples  VI 403 

Inequalities 409 

Limits 417 

Indeterminate  Forms 421 

Exponentials  and  Logarithms 426 

Exponentiation 427 

Loga):ithmic  O^^eration 430 

Natural  Logarithms 436 

Convergency  and  Divergency  of  Series .  447 

Indeterminate  Coefficients 465 

Application  to  Integral  Functions 469 

Application  to  Partial  Fractions 472 

Application  to  Expansion  of  Functions 476 

Application  to  Summation  of  Series 480 

Permutations  and  Combinations 485 

The  Binomial  Theorem  for  an  Integral  Index 496 

The  Binomial  Theorem  :  any  Index 507 

Exponential  and  Logarithmic  Series 514 

Logarithmic  Computation 521 


xu 


CONTENTS. 


CHAPTER 

XXXVII. 
XXXVIII. 


XXXIX. 
XL. 


XLI. 
XLII. 

XLIII. 

XLIV. 

XLV. 

XLVI. 

XL  VII. 

XLVIII. 


PAQK 

Continued.  Fractions 630 

Scales  of  Notation J 548 

Miscellaneous  Examples  VII ,/ 654 

Determinants , 561 

Rational  Factors  and  Higher  Equations 585 

To  Factor  x^  +  y^ -{- z^  -  S  xyzJ 585 

To  Factor  the  Cubic 586 

To  Factor  the  Biquadratic 588 

To  Factor  ax^  +  2  hxy^  -{- by^ +  2  gx  +  2fy  +  c. .  590 

Simultaneous  Equations 594 

Horner's  Method 598 

Roots  and  Coefficients (60&: 

Symmetric  Functions 608  . 

Graphical  Methods 613 

Binomial  Equations 617 

Derivatives 624 

Maxima  and  Minima 631 

Successive  Derivatives 633 

Taylor's  Theorem 636 

Continuity 638 

Rolle's  Theorem 639 

Transformation 641 

Reciprocal  Equations 645 

Criteria  for  Real  Roots 650 

Descartes'  Rule  of  Signs 653 

Sturm's  Theorem 658 

Miscellaneous  Examples  VIII 666 


\  D  B  A  ^  V 


ALGEBRA. 


-»o»{oo- 


CHAPTER  I. 
Introductory  Lessons. 

1.  Signs  and  Symbols  in  Algebra.  In  algebra  a  letter  is 
frequently  used  to  denote  a  number.  When  so  used  in 
a  connected  series  of  operations,  a  given  problem  for 
example,  the  letter  must  be  regarded  as  standing  for  the 
same  number  in  all  the  operations  of  the  series.  Several 
distinct  letters  may  be  used  to  denote  several  distinct 
numbers  in  the  same  problem. 

In  these  Introductory  Lessons  the  signs  +,  — ,  X,  -^, 
and  =,  have  the  usual  significations  given  to  them  in 
arithmetic.  Their  formal  definitions  are  given  in  Arts. 
12,  13,  14,  15,  and  21,  of  the  next  chapter. 

When  letters  are  used  to  represent  numbers,  the  sign 
of  multiplication  is  frequently  omitted.  Thus  ab  means 
a  xb. 

2.  The  Method  of  Algebra.  The  statement  of  an  equality, 
made  by  placing  the  sign  =  between  two  numbers,  or  two 
sets  of  numbers,  indicating  that  the  two  numbers,  or  two 
sets  of  numbers,  are  equal  to  one  another,  is  called  an 
equation. 


%  TNTPvODUCroilY   LESSONS. 

Thus  2  a;  =  5  and  a;  +  2  =  a  +  3 

are  equations.  Problems  stated  in  the  form  of  equations 
are  said  to  be  stated  algebraically. 

In  this  introductory  chapter  some  of  the  uses  of  alge- 
bra will  be  explained  by  stating  and  solving,  with  the 
help  of  equations,  certain  classes  of  problems  already 
familiar  to  the  learner  who  has  studied  arithmetic. 

In  applying  the  method  of  algebra,  the  first  step  is  to 
translate  the  ordinary  language  in  which  the  problem  is 
expressed  into  a  concise  symbolical  statement,  and  the 
form  used  for  this  purpose  is  the  equation.  The  solution 
of  this  equation  then  solves  the  problem  itself.  Algebra 
has  no  one  universal  rule  that  will  put  every  problem 
expressed  in  ordinary  language  into  the  algebraic  form, 
but  rather  it  devises  ways  and  means  for  solving  the 
various  classes  of  equations  that  are  produced  by  dif- 
ferent practical  problems. 

Our  first  concern  is  to  state  the  given  problem  in  the 
form  of  an  equation,  and  the  way  to  do  this  must  be 
sought  in  the  language  of  the  problem  itself.  A  few 
simple  examples  will  best  explain  how. 

Ex.  1.  If  %  10  were  added  to  twice  the  money  I  have,  the  result 
would  be  $  150.     How  much  money  have  I  ? 

The  specifications  of  this  problem  may  be  expressed  more  con- 
cisely thus : 

Twice  the  money  I  have^  added  to  f  10,  equals  $  150,  or  again, 
using  the  signs  +  and  =  in  the  ordinary  arithmetical  sense, 

Twice  the  money  I  have  +  $  10  =  $  150. 

The  problem  is  here  stated  in  the  form  of  an  equation,  which  as 
yet,  however,  is  not  purely  algebraic. 


INTRODUCTORY  LESSONS.  3 

This  analysis  will  not  be  in  any  way  disturbed  by  the 
replacement  of  10  and  150  by  any  other  numbers  we 
choose  to  insert,  and  we  may  analyze  with  equal  facility 
the  following  problem,  in  which  a  and  h  stand  for  any 
numbers,  although  we  impose,  for  the  present,  the  con- 
dition that  h  shall  be  greater  than  a. 

Ex.  2.  If  a  dollars  were  added  to  twice  the  money  I  have,  the 
result  would  be  h  dollars.    How  much  money  have  I  ? 
Stated  in  the  briefer  form,  this  problem  says ; 

Twice  the  number  of  dollars  I  have  +  ^  a  =  $  6. 

Ex.  3.  The  sum  of  two  numbers  is  50,  and  their  difference  is  20. 
What  are  the  numbers  ? 

The  first  condition  of  this  problem  asserts  that 

The  larger  number  +  the  smaller  number  =  50  ; 
and  since  we  obviously  get  the  larger  number  by  adding  their  dif- 
ference to  the  smaller,  the  second  condition  asserts  that 

20  +  the  smaller  number  =  the  larger  number. 
Hence,  replacing  '  the  larger  number '  in  the  former  of  these  two 
statements  by  its  equal  '  20  +  «^e  smaller  number^'  we  obtain 
20  +  the  smaller  number  +  the  smaller  number  =  50, 
or,  in  briefer  form, 

20  +  twice  the  smaller  number^  50. 
From  this  statement  the  smaller  number  is  at  once  seen  to  be  15. 

Observe  that  here,  as  in  Ex.  1,  the  numbers  20  and  50 
may  be  replaced  by  letters  representing  any  other  num- 
bers, without  in  any  way  affecting  the  analysis  of  the 
problem,  or  the  form  of  its  solution. 

These  examples  point  to  the  fact  (they  do  not  prove 
it)  that  no  matter  what  the  relation  of  the  quantities 
involved  in  a  determinate  problem  may  be,  it  can  be 
expressed  in  the  form  of  an  equation. 


4  INTRODUCTORY  LESSONS. 

But  although,  the  abbreviated  statements  thus  far  pro- 
posed are  all  more  concise  than  the  corresponding  state- 
ments in  ordinary  language,  they  can  be  reduced  to  still 
more  convenient  form.  Thus,  in  Ex.  1,  if  we  briefly 
indicate  ^  the  number  of  dollars  I  have '  by  soine  symbol 
or  letter,  as  $  x,  the  equation 

Twice  the  number  of  dollars  I  have  -f-  $  10  =  $  150 

becomes  2x$a;  +  $10  =  $  150, 

and  is  now  -more  strictly  algebraic.  Any  symbol,  for 
example  ($?),  may  here  be  used  to  stand  for  ^the  num- 
ber of  dollars  I  have,^  but  it  is  the  universal  custom  to  use 
a  letter  for  this  purpose,  and  obviously  it  may  be  any 
letter. 

It  must  be  noticed  that,  in  the  algebraic  statement  of 
a  problem,  x  is  an  abstract  number,  which  may  turn  out 
to  be  5,  or  11,  or  |,  or  in  fact  any  number,  and  that, 
therefore,  x  dollars  means  x  times  one  dollar,  or  x  x  {1 
dollar).  Similarly,  x  inches  means  a?  x  (1  inch),  etc.  In 
other  words,  x  dollars  is  a  product,  the  first  factor  of 
which  is  the  abstract  number  x,  the  second  being  the 
concrete  unit  one  dollar,  in  terms  of  which  all  the  quan- 
tities involved  in  the  problem  must  be  expressed.  The 
unit  beinp  uhen  well  understood,  we  drop  it  out  of  the 
equation,    /hich  in  the  case  of  Ex.  1  then  becomes 

2a; +  10  =  150, 
a  pure  algebraic  equation.     The  purely  algebraical  state- 
ment of  Ex.  2  is,  similarly, 

2x-\-a  =  b, 
and  that  of  Ex.  3  is     20  +  2  a;  =  50, 
in  which  x  stands  for  '  smaller  number/ 


INTRODUCTORY   LESSONS.  O 

It  is  important  that  the  beginner  clearly  understand 
what  these  algebraic  equations  mean.  They  can  be 
easily  translated  backwards  into  the  ordinary  language 
in  which  the  problems  were  originally  expressed.  An 
equation  is  thus  simply  a  sentence  in  algebraic  language 
expressing  the  conditions  of  the  problem.  And  besides 
its  conciseness  of  statement,  giving  at  a  glance  the  rela- 
tions between  the  quantities  involved  in  it,  the  theory  of 
algebra  will  eventually  show  that  out  of  it  may  be 
evolved,  by  simple  processes,  the  value  of  the  impor- 
tant quantity  ic,  the  unknown  quantity,  which,  when 
found,  answers  the  question  propounded  by  the  problem. 

State  the  following  problems  algebraically : 

Ex.  4.  Wliat  number  .is  that  whose  half  added  to  73  will  pro- 
duce 85  ? 

Ex.  6.  What  number  is  that  whose  nth  part  added  to  a  will 
produce  b  ? 

Ex.  6.  A  boy  had  a  apples.  He  gave  a  certain  number  of 
them  to  his  companions,  and  then  had  b  times  as  many  as  he  had 
given  away.     How  many  did  he  give  away  ? 

Let  the  learner  now  read  the  axioms  of  Art.  93. 

3.  Problems  involving  the  Four  Fundamental  Processes  of 
Arithmetic.  r  • 

Ex.  1.  If  the  sum  of  x  and  2  x  and  3  a;  be  42,  whlft  number  does 
X  stand  for  ? 

The  algebraic  statement  of  the  problem  is 
x-{-2x  +  3x  =  42; 
and  this  is  the  same  as  6x  =  42  ; 

for,  obviously,  any  number  added  to  twice  itself  and  three  times 
itself  is  six  times  the  number.     Hence,  since  6  times  x.is  42, 

X  =  7. 


6  INTRODUCTORY   LESSONS. 

Ex.  2.  If  5  ic  be  subtracted  from  the  sum  of  3  x,  4  x,  and  x  +  6, 
the  result  is  35.     What  number  does  x  stand  for  ? 

The  algebraic  statement  of  the  problem  is 

Sx-{-4:X  +  x  +  6  —  5x  =  S6. 
In  order  to  find  what  x  stands  for,  we  first  combine  the  four 
terms  containing  x  into  one  term,  which,  by  the  rules  of  addition, 
subtraction,   and  multiplication  in  arithmetic,  obviously  is  3  x ; 
hence  3  x  +  5  =  35. 

From  both  sides  of  this  equation  subtract  5  ;  then 
3x  =  30, 
whence  cc  =  10. 

Ex.  3.  The  sum  of  two  numbers  is  50,  and  their  difference  is  20, 
What  are  the  numbers  ? 

This  is  the  third  problem  of  Art.  2.  It  was  there  stated  alge- 
braically ;  we  now  solve  it  in  the  purely  algebraic  form.  Let  x 
represent  the  smaller  number ;  then  20  +  x  is  the  larger,  and,  by 
the  conditions  of  the  problem, 

20  +  a:  +  «  =  50. 
We  now  subtract  20  from  both  sides  of  the  equation,  and  obtain 

2  X  =  30 ; 
whence  x  =  15,  the  smaller  number, 

and  20  +  X  =  35,  the  larger  number, 

Ex.  4.  The  larger  of  two  numbers  is  3  times  the  smaller,  and 
their  difference  is  92.     What  are  the  numbers  ? 

Let  X  represent  the  smaller  number  ;  then  3  x  is  the  larger,  and 
by  the  second  condition  of  the  problem, 

3  X  -  X  1=  92. 
But  three  times  any  number  less  the  number  itself  is  obviouslj 
twice  the  number.     Hence 

2x  =  92, 
and  X  =  46. 


INTRODUCTORY   LESSONS.  7 

Ex.  5.  A,  B,  and  C  together  have  $75  ;  A  has  twice  as  much  as 
B,  and  B  has  $  5  less  than  C.    How  much  has  each  ? 

Let  X  represent  B's  share ;  then  2  x  is  A's  share,  and  x  +  5  is 
C's  share,  and  the  algebraic  statement  of  the  problem  is, 

a;  +  2x  +  x  +  5  =  75. 

From  this  equation  are  obtained,  by  the  processes  already  explained 
in  the  preceding  examples, 

a;  =  17|,  2ic  =  35,  x  +  5  =  22^. 

Ex.  6.  An  indicated  sum,  as  (n  +  ^n  —  1),  is  frequently  placed 
in  a  parenthesis  in  order  to  indicate  that  an  operation  is  to  be 
performed  upon  the  number  which  the  sum  represents.  Thus,  if 
n  =  6, 

w  +  I  w  -  1  =  8, 

and  2  X  (n  +  ^  w  -  1)  =  2  X  8  =  16. 

This  result  will  again  be  the  same  if  each  term  of  the 
sum  be  multiplied  by  2  before  the  substitution  of  6  for 
n  is  made.  Thus,  whatever  number  n  may  represent,  we 
shall  find  it  to  be  always  true  that 

2x(n  +  iM-l)  =  2xnH-2xin-2xl, 

and  we  verify  this  result  in  the  present  instance  by  put- 
ting 6  in  place  of  n,  thus  obtaining 

2x6  +  2x1x6-2  =  16. 

It  is  of  extreme  importance  to  observe  that  we  have 
here,  not  a  proof  of  a  principle  or  formula,  but  a  verifica- 
tion of  it  in  a  particular  instance- 
Ex.  7.  If  n  stand  for  9,  what  numbers  are  represented  by 
2n,  5m,  n  +  3,  w  -  5,  2w  +  3,  3n-2,  3(n  +  1),  3(w  -  4), 
3  n  +  2  w  -  4  w,  K^  -  3),  3(^  w  -  1)? 


8  INTRODUCTORY  LESSONS. 

Ex.  8.  John  is  4  years  older  than  William,  and  4  times  John's 
age  is  5  times  William's.     What  are  their  respective  ages  ? 

Let  X  represent  John's  age ;  then  ic  —  4  is  William's  age,  and 
the  algebraic  statement  of  the  problem  is 

5(a;  — 4)  =  4  a;. 

Here  the  parentheses  enclosing  a;  —  4  indicate  that  both  x  and  4 
are  to  be  multiplied  by  5.  Performing  this  multiplication,  we 
obtain 

5  X  -  20  =  4  a^ ; 

and  from  this  equation,  by  adding  20  and  subtracting  4  a;,  perform- 
ing the  operations  on  both  sides  of  the  equation,  we  find 

X  =  20,  John's  age, 

and  then  aj  —  4  =  16,  William's  age. 

EXAMPLES  I. 

1.  Divide  15  into  two  parts  whose  difference  is  7. 

2.  The  sum  of  two  numbers  is  a,  and  their  difference  is  6. 
What  are  the  numbers  ? 

3.  A  and  B  together  own  140  acres  of  land,  and  B  owns  3  times 
as  much  as  A.     How  many  acres  has  each  ? 

4.  A  man  having  $92  spent  a  part  of  it,  and  then  had  3  times 
as  much  as  he  had  spent.     How  much  did  he  spend  ? 

5.  A  farmer  sold  15  bushels  of  barley  at  a  certain  price,  and 
25  bushels  of  wheat  at  twice  the  price  of  the  barley,  and  he  received 
$14  more  for  the  wheat  than  for  the  barley.  What  were  the 
prices  per  bushel  of  the  wheat  and  of  the  barley  ? 

6.  Twenty-four  coins,  consisting  of  dimes  and  half-dollars, 
amount  to  %  6.    How  many  coins  of  each  kind  are  there  ? 

7.  A  boy  gave  one-third  of  all  the  apples  he  had  and  one-third 
of  an  apple  more  to  his  sister,  and  then  had  one  apple  left.  How 
many  apples  did  he  give  to  his  sister  ?  ! 


INTRODUCTORY  LESSONS.  9 

8.  What  number  is  that  whose  double  exceeds  its  half  by  27  ? 

9.  Verify  the  following  equations  by  putting  some  number  (any 
number)  in  place  ot  n: 

(1)  |w  +  n  =  fw, 

(2)  3  x(n+l)+l  =  3w  +  4, 

(3)  3  X  (n  -  4)  +  2  =  3  «  -  10, 

(4)  5n-2(w- l)=5?i-2n  +  2 

=  3/1  +  2, 
(6)        J(n- l)+2  =  ^w-i  +  2 

(6)  |n-2(2-^n)  =  Jw-4  +  n 
=  |n-4.    . 

10.  A  can  do  a  piece  of  work  in  10  days,  A  and  B  together  in  6 
days.     In  how  many  days  can  B  do  it  alone  ? 

;  11.)  The  age  of  a  father  is  now  7  times  that  of  his  son,  but  in  3 
years  it  will  be  only  5  times  that  of  his  son.    How  old  are  each  ? 

12.   At  what  time  between  2  and  3  o'clock  are  the  hour  and 
minute  hands  of  a  watch  together  ? 

4.   Problems  involving  the  Eule  of  Three. 

Ex.  1.    What  number  has  to  12  the  same  ratio  as  57  to  9  ? 

Let  X  represent  the  number.     Then  if  the  ratio  of  57  to  9  be 
denoted  by  57  :  9,  the  conditions  of  the  problem  require  that 

X  :  12  =  57  :  9  ; 

or  this  statement  may  be  made  in  the  equivalent  fractional  form 

12      9* 
Multiplying  both  sides  of  this  equation  by  12,  we  obtain 


10  INTRODUCTORY  LESSONS. 

Ex.  2.   How  long  will  it  take  to  fill  a  cistern  of  165  gallons  by 
a  pipe  that  fills  one  of  120  gallons  in  8  minutes  ? 

If  X  represent  the  number  of  minutes  it  takes  to  fill  the  larger 
cistern,  then  by  the  rule  of  three  (proportion) 

a;  ^165 
8~120' 

and  by  multiplication  of  both  sides  of  this  equation  by  8, 
„     8  X  165 


120 


=  11. 


Ex.  3.  A  train  goes  at  uniform  speed  one-third  of  a  mile  in 
20  seconds.    What  is  its  speed  per  hour? 

Let  X  represent  the  number  of  miles  per  hour  which  denotes  the 
speed  of  the  train.  Then,  by  the  rule  of  three,  or  proportion,  the 
algebraic  statement  of  the  problem  is 

3600"  20' 

since  there  are  3600  seconds  in  an  hour.     From  this  equation,  by 
the  proper  multiplications  and  reductions,  we  find  x  to  be 

-        3600   ^g()_ 


3  x20 


Ex.  4.  If  it  require  15  men  to  build  a  house  in  92  days,  how 
many  men  must  be  employed  in  order  to  build  it  in  60  days  ? 

This  problem  may  be  stated  in  the  form  of  a  proportion,  or  as 
follows.  It  requires  15  x  92  days'  work  for  the  building  of  the 
house  ;  hence,  if  x  stand  for  the  number  of  men  that  can  build  it; 
in  60  days, 

X  X  60  =  15  X  92. 

Therefore 

«  =  152192  =  23. 
60 


INTRODUCTORY  LESSONS.  11 

Ex,  5.  If  it  require  a  men  to  build  a  house  in  6  days,  how 
many  men  must  be  employed  in  order  to  build  it  in  c  days  ? 

It  requires  a  x  6  days'  work  for  the  building  of  the  house  ; 
hence,  if  x  stand  for  the  number  of  men  that  can  build  it  in 
c  days, 

xY.c  =  ay.h. 
Therefore 

„     a  X  6 


EXAMPLES  II. 

1.  Divide  $20  into  two  parts  whose  ratio  to  one  another  is 
the  same  as  the  ratio  of  3  to  7. 

2.  If  5  tons  of  hay  cost  $84,  how  many  tons  can  be  bought  for 
$270? 

3.  If  the  speed  of  a  train  be  45  miles  an  hour,  how  long  will  it 
take  it  to  go  360  miles  ? 

4.  How  long  will  it  take  to  fill  a  cistern  containing  a  gallons 
by  a  pipe  that  fills  a  cistern  containing  6  gallons  in  c  minutes  ? 

5.  A  train  goes  at  uniform  speed  a  miles  in  h  minutes.     What 
is  its  speed  per  hour  ? 

6.  It  requires  a  men  to  build  a  house  in  h  days.     How  long 
will  it  take  c  men  to  build  it  ? 

7.  In  Ex.  6  put  15  for  a,  75  for  &,  and  20  for  c,  and  work 
out  the  consequent  numerical  result. 

8.  A's  age  is  now  twice  B's  age  ;  but  in  five  years  their  ages 
will  be  in  the  ratio  of  7  to  4.     What  is  the  age  of  each  ? 

9.  If  the  wages  of  6  men  for  a  days  be  $60,  what  will  be  the 
wages  of  13  men  for  the  same  time  and  at  the  same  rate  ? 

10.    If  7  men  earn  $  105  in  6  days,  how  many  men,  working  for 
the  same  rate  of  wages,  will  earn  $  157.50  in  7  days  ? 


12  INTRODUCTORY  LESSONS. 

11.  If  10  men  can  reap  in  3  days  a  field  whose  length  is  1200 
feet,  and  breadth  800  feet,  what  is  the  breadth  of  a  field  whose 
length  is  1000  feet  which  12  men  can  reap  in  4  days  ? 

5.   Problems  in  Percentage  and  Interest. 

Ex.  1.  Of  a  herd  of  cows,  280  are  Jerseys,  and  these  are  35% 
of  the  entire  herd.     How  many  cows  are  there  in  the  herd  ? 

Let  X  stand  for  the  number  of  cows  in  the  herd.  Then  ^-^-^  x  x 
is  the  number  of  Jerseys.  Hence,  the  algebraic  statement  of  the 
problem  is 

tVo  X  oj  =  280, 

and  the  value  of  ic,  derived  from  this  equation,  is 

x  =  M2i^  =  800. 
35 

Ex.  2.  A  town  lost  7%  of  its  population,  and  then  had  6045 
inhabitants.     What  was  its  population  before  the  loss  ? 

Let  05  be  the  original,  unknown  number  of  inhabitants.  Then 
Y^^  X  X  was  the  loss,  and  x  —  j^  x  x  was  the  number  of  inhabi- 
tants remaining  after  the  loss.     Hence 

X-  ^\^Y.x-  6045. 

Now  X  less  seven  hundredths  of  x  is  ninety-three  hundredths  of  x ; 

therefore  tV^  x  x  =  6045, 

,  ^  100    X    6045  r,.r.r. 

and  X  — — =  6500. 

93 

Ex.  3.  The  amount  due  on  a  note  for  $800  at  5%  simple  in- 
terest was  %  900.     For  what  length  of  time  was  interest  reckoned  ? 

The  interest  for  one  year  is  yf^  x  800  dollars,  or  $40;  hence 
the  interest  for  x  years  is  40  x  cc  dollars,  and  the  amount  due  after 
X  years  is  800  +  40  x  cc  dollars,  and  this,  by  the  conditions  of  the 
problem,  is  $900.  The  equation  expressing  the  conditions  there- 
fore is 

800 +  40  a;  =  900. 


INTEODUCTORY  LESSONS.  13 

Subtract  800,  and  divide  by  40,  performing  these  operations  on 
both  sides  of  the  equation  ;  then 

900-800_oi 
40  ^ 

Thus  the  length  of  time  for  which  interest  was  reckoned  is  2^ 
years. 

Ex.4.  What  principal  amounts  to  $1456  in  2  years  at  6% 
simple  interest  ? 

If  X  be  the  principal,  ■^%^  x  a;  is  the  interest  for  one  year,  twice 
this  is  the  interest  for  two  years,  and  x  4-  (2  x  ^^-^  x  x)  is  the 
amount.     Hence 


V 100     ; 


the  parenthesis  being  here  used  to  indicate  that  the  entire  product 
within  it  is  to  be  added  to  x.     (See  Exs.  6,  7,  8,  Art.  3.) 
But  y\^  of  X  added  to  x  is  |^§  of  x.     Therefore 

IH  X  X  =  1456, 
and  ^^100x1456^^3^ 

112 

The  required  principal  is  $1300. 

Ex.  6.  The  amount  due,  after  Z\  years,  on  a  note  for  $1500 
bearing  simple  interest,  was  $1788.75.  'What  was  the  rate  of 
interest  ? 

Let  X  stand  for  the  rate  of  interest ;  then  1500  x  x  is  the  interest 
for  one  year,  3|  x  1500  x  x,  or  5250  x  x,  is  the  interest  for  3^ 
years,  and  1500  +  (5250  x  x)  is  the  amount  due.    Hence 

1500 +  (5250  X  x)  =  1788.75. 

Subtract  1500,  and  divide  by  5250,  performing  these  operations  on 
both  sides  of  the  equation  ;  then 

^^1788.75-1500^  Qgg^ 
5250 

Thus  the  rate  of  interest  was  5^  per  cent. 


14  INTRODUCTOKY   LESSONS. 

EXAMPLES   III. 

1.  What  number  increased  by  |  of  25  %  of  itself  equals  315  ? 

2.  A  city  gained  13  %  in  population,  and  then  had  80,456  inhabi- 
tants.    What  was  its  population  before  the  gain  ? 

3.  The  population  of  a  city  increased  from  31,000  to  33,945 
in  ten  years.  What  was  the  average  per  cent  of  increase  per 
annum  ? 

4.  The  annual  rent  of  a  house  is  $240,  and  this  is  8%  of  its 
value.     What  is  its  value  ? 

5.  What  sum  bearing  interest  at  5i%  will  yield  an  annual 
income  of  f  1500  ? 

6.  In  what  time  will  any  sum  of  money  double  itself  at  8% 
simple  interest  ? 

7.  The  amount  due,  after  a  years,  on  a  note  for  h  dollars  bear- 
ing simple  interest,  was  c  dollars.     What  was  the  rate  of  interest  ? 

6.  Problems  involving  the  Use  of  Negative  Numbers.  In 
the  second  problem  of  Art.  2,  whose  statement  in  the 
algebraic  form  was  found  to  be 

2  a;  +  a  =  6, 

and  whose  solution  is  therefore 

a;  =  1(6 -a), 

we  said  that  h  must  be  greater  than  a.  Now  while  this 
restriction  is  necessary  in  arithmetic,  the  method  of 
algebra  obtains  a  solution  for  all  values  of  a  and  h.  In 
accordance  with  the  enlarged  algebraic  view,  the  problem 
under  consideration  may  be  stated  as  follows : 

Ex.  1.  I  have  no  money ;  but  balance  my  accounts  by  setting 
off  debits  against  credits,  then  double  the  thus  ascertained  value 


INTRODUCTORY  LESSONS.  15 

of  my  possessions,  and  add  a  dollars  to  it,  and  the  result  will  be  h 
dollars.  Can  I  pay  my  debts  with  the  money  due  me,  and  how 
much  money,  if  any,  shall  I  have  after  all  accounts  are  settled  ? 

If  the  excess  of  credits  over  debits  be  x  dollars,  then  by  the  con- 
ditions of  the  problem  2  x  +  a  dollars  is  equal  to  h  dollars,  or,  in 
the  purely  algebraic  form, 

2x  +  a  =  h. 

From  this  equation  we  obtain,  as  previously, 

Thus,  when  all  accounts  are  settled,  I  shall  have  \  {h  —  a)  dollars. 
Let  us  see  what  will  happen  under  the  three  different  suppositions 
that  may  here  be  made. 

(i.)  If  the  credits  exceed  the  debits  (in  amount),  I  shall  be 
able  to  pay  my  creditors  with  the  money  due  me  and  have  some- 
thing left.  This  something  is  \  (h  —  a)  dollars,  and  obviously  h 
must  be  greater  than  a. 

(ii.)  If  the  credits  be  just  equal  to  the  debits,  they  cancel  each 
other  in  the  final  settlement  of  accounts,  and  I  shall  have  just 
nothing  left.  This  nothing  is  again  \  (fi  —  a)  dollars,  and  in  order 
that  I  (6  —  a)  may  be  zero,  h  must  be  equal  to  a. 

(iii.)  But  if  the  debits  exceed  the  credits,  I  shall  fail  to  cancel 
all  my  indebtedness  with  the  money  due  me,  and  shall  have  some 
debts  remaining  and  no  money  to  pay  them  with.  I  no  longer 
have  possessions,  but  debts.  The  amount  of  my  indebtedness  is 
exactly  \  (a  —  6),  and  the  condition  for  this  state  of  affairs  is 
that  a  shall  be  greater  than  b. 

Now  algebra  takes  account  of  this  condition  by  the  simple  device 
of  writing  the  minus  sign  before  the  positive  number  ^  (a  —  6) , 
and  calling  the  result  negative  ;  thus 

ic  =  1  (6  —  a)  =  —  ^  (a  —  6), 

a  negative  number,  representing  indebtedness,  if  a  be  greater  than 
b.    The  number  J  (Jb  —  a)  is  therefore  the  true  answer  to  the  ques- 
tion propounded  by  the  problem,  under  all  circumstances. 
Consider  a  concrete  illustration : 


16  INTRODUCTORY   LESSONS. 

Ex.  2.  Restate  the  problem  with  55  in  place  of  a,  and  5  in  place 
of  b.    The  solution  then  is 

a;  =  1  (5  _  55)  =  _  1  (55  -  5)  =  -25  ; 

and  the  interpretation  of  this  negative  result  is  that  the  debits 
exceed  the  credits  by  $  25.  Thus,  if  credits  =  $  100,  then  debits 
=  I  125  ;  or  if  credits  =  $  5000,  then  debits  =  $  5025,  etc. 

We  may  verify  the  conclusion  here  reached  by  performing  the 
operations  indicated  in  the  problem  upon  —  25.  We  first  double 
—  25  and  obtain  —  50,  then  to  this  negative  number  —  50  we  add 
the  positive  number  55  and  obtain  5.  These  are  some  of  the  ordi- 
nary processes  of  algebra  whose  meaning  will  be  more  fully  ex- 
plained in  the  next  two  chapters. 

Ex.  3.  A  steam  launch,  whose  speed  in  still  water  is  a  feet  per 
minute,  goes  up  an  estuary  a  distance  of  b  feet  in  c  minutes.  Show 
that  the  tide  is  running  at  the  rate  of  (6  h-  c)  —  a  feet  per  minute,  — 
inward  if  6  -=-  c  be  greater  than  a,  outward  if  6  -r-  c  be  less  than  a. 

7.  The  foregoing  problems  have  brought  to  view  some 
of  the  purposes  of  algebra  and  have  shown  how,  by  its 
larger  methods,  some  of  the  inconvenient  limitations  to 
which  arithmetic  is  necessarily  subject  may  be  removed. 
Other  like  problems,  to  be  considered  in  subsequent 
chapters,  will  still  further  attest  the  superiority  of  these 
methods.  But  since  many  of  them  cannot  be  success- 
fully attempted  until  the  fundamental  operations  of 
algebra  are  learned,  our  next  task  must  be  a  study  of  the 
fundamental  operations  of  algebra. 


LANGUAGE  OF  ALGEBRA.      DEFINITIONS.         17 


CHAPTER  II. 
The  Language  of  Algebra.    Definitions. 

8.  Algebra  is  that  branch  of  mathematics  which  treats 
of  the  relations  of  numbers  as  expressed  in  equations  and 
in  what  are  known  as  algebraic  forms,  or  expressions. 

These  algebraic  equations  and  expressions  constitute 
the  language  of  algebra. 

9.  The  Operations  of  Algebra*  are  conveniently  grouped 
in  four  main  divisions,  each  of  which  comprises  two  con- 
trasted kinds  of  operation,  called  respectively  a  direct 
process  and  an  inverse  process.  These  eight  processes, 
thus  classified,  are : 

Direct  Processes  :  Inverse  Processes  : 

I.     1.  Addition,  2.  Subtraction. 

II.     3.  Multiplication,  4.  Division. 

III.     5.  Involution,  6.  Evolution. 

^IV.     7.  Exponentiation,  8.  Logarithmic  Operation. 

The  direct  process  having  been  defined,  its  inverse  is 
described  as  that  operation  which  annuls,  or  undoes,  the 
direct  process. 

*  Another  point  of  view  enables  us  to  regard  all  the  operations  of 
algebra  as  forms  of  the  addition  process.  But  the  investigations  neces- 
sary to  establish  this  conclusion  are  not  appropriate  in  a  merely  ele- 
mentary treatise. 

B 


18         LANGUAGE   OF   ALGEBRA.      DEFINITIONS. 

A  full  explanation  of  the  character  of  these  processes, 
as  fundamental  parts  of  algebraic  science,  forms  a  neces- 
sary introduction  to  the  problems  that  algebra  under- 
takes to  solve.  They  will  be  discussed  in  the  order  here 
indicated,  the  first  four  in  Chapters  III.,  IV.,  and  V.,  the 
others  somewhat  later,  as  occasion  for  their  use  arises. 

10.  The  Symbols  of  Algebra.  For  the  convenient  rep- 
resentation of  these  operations  an  algebraic  symbolism 
has  been  devised;  and,  as  has  been  exemplified  in  the 
problems  of  the  Introductory  Lessons,  the  signs  used  in 
arithmetic  are  adopted  as  a  part  of  it.  »"'But  for  the 
adequate  representation  of  all  its  operations,  algebra 
requires  an  enlarged  symbolism  of  which  arithmetic 
needs  only  a  part. 

The  symbols  of  elementary  algebra  are  of  five  kinds  : 

(i.)  Symbols  of  Quantity,  usually  letters  of  the  alpha- 
bet, employed  to  represent  numbers. 

(ii.)  Symbols  of  Operation,  usually  called  the  signs  of 
algebra,  employed  to  indicate  processes  to  be  applied  to 
the  symbols  of  number.     (  + ,  — ,  ~,  x,  ^,  %  -y/,  log.) 

(iii.)  Symbols  of  Relation,  by  means  of  which  compari- 
sons of  numbers  are  expressed.     (  =  ,  =,  >,  <.) 

(iv.)  Abbreviations  of  frequently  recurring  words  or 
phrases.     (.*.,  •••,  •••.)^ 

(v.)  Symbols  of  Aggregation.  (Parentheses,  braces, 
square  brackets,  the  vinculum.) 

SYMBOLS   OF   QUANTITY. 

11.  In  arithmetic,  numbers  are  represented  by  figures, 
each  of  which  has  one,  and  only  one,  meaning. 


LANGUAGE  OF   ALGEBRA.      DEFINITIONS.  19 

^  In  algebra,  numbers  are  represented  either  by  figures 
or  by  the  letters  of  the  alphabet. 

In  an  arithmetical  product,  such  as  3  x  5,  it  is  obvious  that  the 
factors  3  and  5  may  be  interchanged,  and  this  law  of  commutation, 
as  it  is  called,  is  easily  proved  to  obtain  for  any  two  numbers. 
But  the  statement  of  the  law  in  the  form  3  x  5  =  5  x  3  is  the  state- 
ment of  a  particular  case  only.  Now  algebra  states  it  in  the  form 
a  X  6  =  6  X  a,  in  which  a  and  6  represent  any  two  numbers  what- 
ever. /The  gain  in  conciseness  and  generality  of  statement  by  the 
use  of  letters' is  thus  apparent. 

The  word  quantity  is  used  in  algebra  as  synonymous 
with  number. 

SYMBOLS  OF   OPERATION. 

12.  The  sign  -f-,  which  is  read  ^plus,^  is  placed  before 
a  number,  or  a  number  represented  by  a  letter,  to  indicate 
that  it  is  to  be  added  to  what  has  gone  before. 

Thus  6  +  3  means  that  3  is  to  be  added  to  6  ;  6  +  3  +  2  means 
that  3  is  to  be  added  to  G  and  then  2  added  to  the  result :  so  also 
a+  b  means  that  the  number  which  is  represented  by  b  is  to  be 
added  to  the  number  which  is  represented  by  a ;  or,  expressed 
more  briefly,  it  means  that  b  is  to  be  added  to  a. 

13.  The  sign  — ,  which  is  read  ^  minus,'  is  placed 
before  a  number  to  indicate  that  it  is  to  be  subtracted 
from  what  has  gone  before. 

Thus  6  —  3  means  that  3  is  to  be  subtracted  from  6,  a  ~  b  means 
that  b  is  to  be  subtracted  from  a,  and  a  —  b  +  c  means  that  b  is  to 
be  subtracted  from  a  and  then  c  added  to  the  result. 

It  should  be  noticed  that  in  a  series  of  additions  and 
subtractions  the  order  of  the  operations  is  from  left  to  right. 
The  meaning  of  the  sign  ~  is  explained  in  Art.  46. 


20         LANGUAGE  OF  ALGEBRA.      DEFINITIONS. 

14.  The  sign  of  multiplication  is  x,  which  is  read 
^multipHed  by'  or  ^into.' 

Thus  6x3  means  that  6  is  to  be  multiplied  by  3,  a  x  6  means 
that  a  is  to  be  multiplied  by  &,  and  a  x  6  x  c  means  that  a  is  to  be 
multiplied  by  h  and  then  the  result  multiplied  by  c. 

The  sign  of  multiplication  is  generally  omitted  between 
two  letters,  or  between  a  number  and  a  letter,  and  the 
letters  are  simply  placed  side  by  side.  Sometimes  the  x 
is  replaced  by  a  point. 

Thus  ah  OT  a-h  means  the  same  as  a  x  6,  and  2  ahc  or  2a'b'C 
the  same  as  2  x  a  x  6  x  c. 

15.  The  sign  of  division  is  -i-,  which  is  read  'divided 
by'  or  'by.' 

Thus  6-7-3  means  that  6  is  to  be  divided  by  3,  a  -4-  6  means  that 
a  is  to  be  divided  by  b,  and  a  -^  b  -^  c  means  that  a  is  to  be  divided 
by  6  and  then  the  result  divided  by  c  ;  also  a  -^  b  x  c  means  that  a 
is  to  be  divided  by  b  and  then  the  result  multiplied  by  c. 

The  operation  of  division  is  often  indicated  by  placing 
the  dividend  over  the  divisor  with  a  line  between  them, 
or  by  separating  the  dividend   from  the  divisor  by  an 

oblique  line  called  the  solidus ;  thus  either  -  or  a/b  is 
frequently  used  instead  of  a  -r-b. 

It  should  be  noticed  that  in  a  series  of  multiplications 
and  divisions  the  order  of  the  operations  is  from  left  to 
right. 

16.  When  two  or  more  numbers  are  multiplied  together, 
the  result  is  called  the  continued  product,  or  simply  the 
product ;  and  each  number  is  called  a  factor  of  the  product. 


LANGUAGE  OF   ALGEBEA.      DEFINITIONS.  21 

17.  When  the  factors  of  a  product  are  considered  as 
divided  into  two  sets,  each  is  called  the  coefficient,  that  is 
the  co-factor,  of  the  other. 

Thus  in  3  ahx,  3  is  the  coefficient  of  ahx  ;  also  3  a  is  the  coeffi- 
cient of  6aj,  and  3  ah  is  the  coefficient  of  x. 

When  one  of  the  factors  of  a  product  is  a  number 
expressed  in  figures,  it  is  called  the  numerical  coefficient 
of  the  other  part  of  the  product. 

EXAMPLES  IV. 
Calculate  the  values  of 

1.  7  +  6  +  4.  3.   il  +  7-12-6.  6.   5 -- 3  x  4. 

2.  6  -  3  +  4.  4.    7  X  6  X  4.  6.    11  x  7  --  12  ^  6. 
If  a  =  1,  6  =  2,  c  =  3,  and  (Z  =  4,  find  the  numerical  values  of 

7.  c-6.  12.    13a-6& +  7c-6d. 

8.  d-a.  13.    18ft -3c-4d  +  9a. 

9.  7a -36.  14.   20a6-3cd. 

10.  106 -6c.  15.  4<?a-26c. 

11.  6a  -  2  6  +  6 c  -  4 d  16.   a6c  +  bed  -\-  cda -{■  dab. 

If  a  =  6,  6  =  2,  c  =  5,  and  d  =  0,  find  the  values  of 

17.  3ac  +  2  6c  +  ca.  19.    a  x  c  ^  6.  21.   2c  -^  a  -r-  6. 

18.  7aKZ  + 96c- 3ca.  20.   a  ^  c  x  6.  22.   3ac-^6. 

23.  What  are  the  coefficients  of  x  in  3a;,  4  6a;,  5  bcx,  and  16  a6ca;  ? 

24.  What  are  the  coefficients  of  xy  in  4xy,  6axy,  7  abxy,  and 
19  abcxy  ?    What  are  the  numerical  coefficients  ? 

'    18.   When  a  product  consists  of  the  same  factor  re- 
peated any  number  of  times,  it  is  called  a  power  of  that 


22         LANGUAGE   OF   ALGEBRA.      DEFINITIONS. 

factor.  Thus  aa  is  called  the  second  power  of  a,  aaa  is 
called  the  third  power  of  a,  aaaa  is  called  the  fourth  power 
of  a,  and  so  on.  Sometimes  a  is  called  the^rs^  poiver  of  a. 
Special  names  are  also  given  to  aa  and  aaa ;  they  are 
called  respectively  the  square  and  the  cube  of  a. 

19.  Instead  of  writing  aa,  aaa,  etc.,  a  more  convenient 
notation  is  adopted  as  follows  :  a^  is  used  instead  of  aa, 
a^  is  used  instead  of  aaa,  and  a"  is  used  instead  of  aaa  •••, 
the  factor  a  being  taken  n  times ;  the  small  figure,  or 
letter,  placed  above  and  to  the  right  of  a,  showing  the 
number  of  times  the  factor  a  is  to  be  taken.  So  also  a^b^ 
is  written  instead  of  aaabb,  and  siniilarly  in  other,  cases. 

The  small  figure,  or  letter,  placed  above  a  symbol  to 
indicate  the  number  of  times  that  symbol  is  to  be  taken 
as  a  factor,  is  called  the  index  or  the  exponent.  Thus,  a" 
means  that  the  factor  a  is  to  be  taken  n  times,  or  that 
the  nth.  power  of  a  is  to  be  taken,  and  n  is  called  the 
index. 

When  the  factor  a  is  only  taken  once,  we  do  not  write 
it  a^,  but  simply  a. 

20.  The  quantity  which  when  squared  is  equal  to  any 
number  a  is  called  the  square  root  of  a,  and  is  represented 
by  the  symbol  -^a,  or  more  often  by  -^a :  thus  2  is  ■y/4:, 
for  22  is  4. 

The  quantity  which  when  cubed  is  equal  to  any  num- 
ber a,  is  called  the  cube  root  of  a,  and  is  represented  by 
the  symbol  -^a :  thus  3  =  ^27,  for  3^  =  27. 
/    In  general,  the  quantity  which  when  raised  to  the  nth 
power,  where  n  is  any  whole  number,  is  equal  to  any 


LANGUAGE   OF   ALGEBRA.      DEFINITIONS.  23 

number  a,  is  called  the  nth.  root  of  a,  and  is  represented 
by  the  symbol  -y/a. 

/  The  sign  ^  was  originally  the  initial  letter  of  the  word 
radix.  It  is  often  called  the  radical  sign. 
/  A  root  which  cannot  be  obtained  exactly  is  called  a 
surd,  or  an  irrational  quantity  :  thus  -^7  and  ^4  are  surds. 
The  approximate  value  of  a  surd,  for  example  of  -y/7, 
can  be  found,  to  any  degree  of  accuracy  which  may  be 
desired,  by  the  ordinary  arithmetical  process ;  but  we  are 
not  required  to  find  these  approximate  values  in  algebra : 
for  us  -^7  is  simply  that  quantity  which  when  squared 
will  become  7. 

SYMBOLS  OF   RELATION. 

21.  The  sign  =,  which  is  read  ^equals,'  or  'is  equal  to/ 
is  placed  between  two  quantities  to  indicate  that  they  are 
equal  to  one  another. 

Thus  6  4-  7  =  12,  which  is  read  five  plus  seven  equals  twelve. 

The  sign  >  indicates  that  the  number  which  precedes 
the  sign  is  greater  than  that  which  follows  it. 

Thus  a>b  means  that  a  is  greater  than  b. 

The  sign  <  indicates  that  the  number  which  precedes 
the  sign  is  less  than  that  which  follows  it. 

Thus  «  <  6  means  that  a  is  less  than  b. 

The  meaning  of  the  sign  =  is  explained  in  Art.  90. 

SYMBOLS   OF   ABBREVIATION. 

22.  The  sign  •.  •  is  written  for  the  word  because  or  since. 
The  sign  .-.  is  written  for  the  word  therefore  or  hence. 


24         LANGUAGE  OF    ALGEBRA.      DEFINITIONS. 

■     The  sign  of  continuation,  .••,  is  used  for  the  words  and 
so  on,  or  and  so  on  to. 

Thus  1, 2,  3,  4,  •••  means  1,  2,  3,  4,  and  so  on,  and  xi,  x<i^  aJs,  •••  Xn 
means  X\^  X2,  Xs^  and  so  on  to  x„. 

ALGEBRAIC   EXPRESSIONS. 

23.    A  collection  of  algebraic  symbols,  that  is  of  letters, 
figures,  and  signs,  is  called  an  algebraical  expression. 
^  The  parts  of  an  algebraical  expression  which  are  con- 
nected by  the  signs  -f-  or  —  are  called  the  terms. 

Thus  2a  —  3bx-\-5cy^  is  an  algebraical  expression 
containing  the   three   terms   2  a,  —  3  bx,  and  -f  5  cy^. 

24. ''When  two  terms  contain  the  same  letters,  every 
one  of  which  is  raised  to  the  same  power  in  both,  they 
are  called  Hke  terms. 

Thus  3  abH^  and  7  ab^x^  are  like  terms  ;  hut  although  3  a^bx^  and 
7  a^b^x^  contain  the  same  letters,  they  are  not  like  terms,  for  all  the 
letters  are  not  raised  to  the  same  power. 

25y  A  monomial  expression  is  one  which  contains  only 
one  term,  and  a  multinomial  expression  is  one  which  con- 
tains more  than  one  term. 

Thus  5  ab^cx  is  a  monomial  expression,  and  a  ■{■  b  is  a  multino- 
mial expression. 

An  expression  which  consists  of  two  terms  is  often 
called  a  binomial  expression,  and  an  expression  which  con- 
sists of  three  terms  is  called  a  trinomial  expression. 

Monomial  and  multinomial  expressions  are  sometimes 
called  respectively  simple  and  compound  expressions. 


LANGUAGE  OF   ALGEBRA.      DEFINITIONS.         25 

SYMBOLS    OF    AGGREGATION. 

26.  The  parentheses  (  ),  the  square  brackets  [  ],  and 
the  braces  \  \ ,  are  used  to  indicate  that  all  operations 
denoted  by  signs  within  the  enclosure  are  to  be  consum- 
mated before  any  operation  without  it  is  performed. 

Thus  (a  +  6)  c  means  that  b  is  to  be  added  to  a  and  that  the 
result  is  to  be  multiplied  by  c ;  again,  (a  +  by  means  that  b  is  to 
be  added  to  a,  and  that  the  cube  of  the  result  is  to  be  formed  ;  also 
(a  -\-  2b)(c  —  Sd)  means  that  2  6  is  to  be  added  to  a  and  that 
3  d  is  to  be  taken  from  c,  and  that  the  first  of  these  results  is  then 
to  be  multiplied  by  the  second. 

A  line  called  a  vinculum  is  often  drawn  over  the  expres- 
sion which  is  to  be  treated  as  a  whole :  thus  a  —  b  —  c  is 
equivalent  to  a  —  {b  —  c)j  and  Va -f  6  is  equivalent  to 

When  no  vinculum  or  bracket  is  used,  a  radical  sign  refers  only 
to  the  number  or  letter  which  immediately  follows  it :  thus  ^^2  a 
means  that  the  square  root  of  2  is  to  be  multiplied  by  a,  whereas 
\/2a  means  the  square  root  of  2 a  ;  also  y/a  +  x  means  that  x  is  to 
be  added  to  the  square  root  of  a,  but  Va  +  x  means  the  square 
root  of  the  sum  of  a  and  x. 

'  The  line  between  the  numerator  and  denominator  of 

a  fraction  acts  as  a  vinculum,  for       '      is  the  same  as 

'  12 

Note. — It  should  be  carefully  noticed  that  every  term  of  an 
algebraical  expression  must  be  added  or  subtracted  as  a  whole, 
as  if  it  were  enclosed  in  brackets.  Thus,  in  the  expression 
a-hbc  —  d^e-\-f,  b  must  be  multiplied  by  c  before  addition,  and 
d  must  be  divided  by  e  before  subtraction,  just  as  if  the  expression 
were  written  a  +  (be)  —  (<?  -^■  e)  +  /. 


26         LANGUAGE   OF   ALGEBRA.      DEFINITIONS. 

Ex.   Find  the  values  of 

(i.)  2a-6c  +  cd-66  -a  +  ?-?,     (ii.)  (a  +  6)^(26  -  3c)2, 

a 
(iii.)  a  +  ftc  +  c«,  (iv.)  ^/{7  a^^-(h  +  c)^  +#}, 

when  a  =  4,  &  =  3,  c  =  1,  c?  =  0.     The  results  are  : 

(i.)  2a-hc  +  cd-Qh-^a  +  — 

a 

=  2x4-3x1  +  1  X  0-6x3 --4  +  ?-^ 

4 

=  8  -  3  +  0  -  I  +  1-  =  1. 

(ii.)  (a  +  &)3(2  6  -  3c)2  =  (4  +  3)3(2  X  3  -  3  x  1)2  =  73  X  32 

=  343  X  9  =  3087. 
(iii.)  a^  +  ftc  _^  c«  =  43  4-  31  +  1*  =  64^+  3  +  1  =  68. 
(iv.)  ^{7a3  +  (&  +  c)3  +  #}  =  ^{7  X  43  +  43  +  03} 

=  ^{7  X  64 +  64  +  0}  =  ^(448 +  64)  =  ^512  =  8. 

EXAMPLES    V. 

1.  Write  down  the  values  of  2*,  33,  43,  4*,  ^64,  ^64,  ^16,  ^125, 
^625,  and  ^32. 

Ifa  =  2,  6  =  3,  c  =  4,  and  <Z  =  5,  find  the  numerical  values  of 

2.  a2  +  62.  4.   6a2-2  62.  e.   a2&2  +  02^2. 

3.  c2  +  c22.  5.   4  62  _  c2  +  5  <Z2.  7.   a62c2  _  ^25^. 
Ifa  =  2,6  =  3,  c  =  4,  and  (2  =  5,  find  the  numericar  values  of 

8.—-^  +  ^-     ^.  \hc  +  \ca  +  lah.     10.  10 c3#  _  ig ^252. 
5       3        27 

11.   Ia262c3_ia62c2^.  12.   ab^_g^b^. 

^  ^  16  20 

If  a  =  5,  6  =  3,  c  =  1,  d  —  0,  find  the  numerical  values  of 

•    13.    (2a  +  56)(36-6c).  14.    (a  +  26)(c  +  2d). 

15.  (3a -46)3- 2(36  -  6c)2  +  2(a(2  +  6c)2. 

16.  4a3  +  463  +  4c3- 3(6 +  c)(c  +  a)(a  + 6). 


LANGUAGE   OF   ALGEBRA.      DEFINITIONS.  27 

17.  5(a  +  c)3(6-c)2- J3(a-2d)3(&  +  3c)2. 

18.  Show  that  x^  —  5  x  +  6  is  equal  to  zero,  if  x  =  2  or  if  x  =  3. 

19.  Show  that  2  x^  -  11  x^  +  17  x  —  6  is  equal  to  zero,  if  x  =  2, 
or  if  x  =  3,  or  if  X  =  2' 

If  a  =  5,  6  =  4,  c  =  ^,  find  the  numerical  values  of 

20.  Va2  -  &2.       21.  V6a.      22.   V26c  +  3a.      23.   y/bc  +  a^. 
24.    ^(2a2  +  62_8c2).  25.  ^(4a2  _  ^52  _,.  jc2  -  1). 

26.  ^(a+b)^(iSab  +  2bc). 

27.  Find  the  numerical  value  of 

(a  +  6)2  (X  +  y)2  -  4  (ax  +  6y) (bx  +  ay), 
if  X  =  5,  ?/  =  8,  a  =  6,  and  6  =  4. 


28.  Find  the  value  of  vs(s  —  rt)(s  —  6)(«  —  c),  when  <z  =  9, 
6  =  12,  c  =  15,  and  s  =  18. 

29.  Find  the  value  of   J^       {7,         s  when  a  =  9,  6  =  12, 

\(s-6)(s-c) 

c  =  15,  and  2s  =a  +  6  +  c. 

30.  Find,  when  a=:8,  6  =  5,  c  =  3,  the  numerical  value  of 

V{2  62c2  +  2  c2a2  +  2  a262  _  a*  _  54  _  c*}. 

31.  Verify  that  a^  —  b^  and  (a  +  6)(a  —  6)  are  equal  to  one 
another  (i.)  when  a  =  6,  6  =  3  ;  (ii.)  when  a  =  9,  6  =  4  ;  and  (iii.) 
when  a  =  12,  6  =  7. 

32.  Verify  that  the  expressions 

a^  -  63,  (a  -  6)  (a^  +  ab  +  62),  (a  -6)8  +  3  ab  (a-b), 
and  (a  +6)3-3  a6(a  +  6)  -  2  63 

are  all  equal  to  one  another  (i.)  when  a  =  3,  6  =  2  ;    (ii.)  when 
a  =  6,  6  =  3  ;  and  (iii.)  when  a  =  5,  6  =  2. 

POSITIVE  AND  NEGATIVE    QUANTITIES. 

27.  All  concrete  quantities  must  be  measured  by  the 
number  of  times  each  contains  some  unit  of  its  own  kind. 


28         LANGUAGE   OF   ALGEBRA.      DEFINITIONS. 

Now  a  sum  of  money  may  be  either  a  receipt  or  a  pay- 
ment, it  may  be  either  a  gain  or  a  Zoss;  motion  along  a 
straight  line  may  be  in  either  of  two  opposite  directions ; 
a  period  of  time  may  be  either  before  or  after  some 
particular  epoch;  and  so  in  very  many  other  cases. 
Thus  many  concrete  magnitudes  are  capable  of  existing 
in  two  diametrically  opposite  states. 

28.  Whatever  kind  of  quantity  we  are  considering, 
4-4  will  stand  for  what  increases  that  quantity  by  4 
units,  and  —4  will  stand  for  whatever  decreases  the 
quantity  by  4  units. 

If  we  are  calculating  the  amount  of  a  man's  property 
(estimated  in  dollars),  4-  4  will  stand  for  what  increases 
his  property  by  $  4,  that  is  +  4  will  stand  for  $  4  that 
he  possesses,  or  that  is  owing  to  him ;  so  also  —  4  will 
stand  for  whatever  decreases  his  property  by  ^4,  that 
is  —  4  will  stand  for  $  4  that  he  owes.  If,  on  the  other 
hand,  we  are  calculating  the  amount  of  a  man's  debts, 
4-  4  will  stand  for  whatever  increases  his  debts,  that  is 
4-  4  will  now  stand  for  a  debt  of  $  4 ;  so  also  —  4  will 
now  stand  for  whatever  decreases  his  debts,  that  is  —  4 
will  stand  for  f  4  that  he  has,  or  that  is  owing  to  him. 

If  we  are  considering  the  amount  of  a  man's  gains, 
4-  4  will  stand  for  what  increases  his  total  gain,  that  is 
4-  4  will  stand  for  a  gain  of  4 ;  so  also  —  4  will  stand 
for  what  decreases  his  total  gain,  that  is  —  4  will  stand 
for  a  loss  of  4.  If,  however,  we  are  calculating  the 
amount  of  a  man's  losses,  4-  4  will  stand  for  a  loss  of  4, 
and  —  4  will  stand  for  a  gain  of  4. 

Again,  if  the  magnitude  to  be  increased  or  diminished 
is  the  distance  from  any  particular  point,  measured  in 


LANGUAGE   OF  ALGEBRA.      DEFINITIONS.  29 

any  particular  direction,  +4  will  stand  for  a  distance 
of  4  units  in  that  direction,  and  —4  will  stand  for  a 
distance  of  4  units  in  the  opposite  direction. 

29.  From  the  above  examples  it  will  be  seen  that  the 
signs  +  and  —  will  serve  to  distinguish  between  magni- 
tudes of  opposite  kinds.  Thus  whatever  +  4  may  rep- 
resent, —  4  will  represent  an  equal  magnitude,  but  of  the 
opposite  kind.  /The  signs  +  and  —  are  therefore  used 
in  algebra  with  two  entirely  different  meanings.  '  In 
addition  to  their  original  meanings  as  signs  of  the  opera- 
tions of  addition  and  subtraction  respectively,  they  are 
also  used  as  marks  of  distinction  between  magnitudes  of 
opposite  kinds./ 

30.  A  quantity  to  which  the  sign  -f  is  prefixed  is 
called  a  positive  quantity,  and  a  quantity  to  which  the 
sign  —  is  prefixed  is  called  a  negative  quantity.  ^ 

The  signs  +  and  —  are  called  respectively  the  positive 
and  negative  signs. 

31.  The  signs  +  and  —  are  often  called  signs  of  affection 
when  they  are  used  to  indicate  a  quality  of  the  quantities 
before  whose  symbols  they  are  placed. 

The  sign  +,  as  a  sign  of  affection,  is  frequently 
omitted;  and  when  neither  the  -f  nor  the  —  sign  is 
prefixed  to  a  term,  tjie  +  sign  is  to  be  understood. 

32.  The  range  of  positive  and  negative  algebraic 
numbers  is  obviously  double  that  of  the  numbers  which 
belong  to  arithmetic.  Thus,  the  series  of  integers  in 
arithmetic  is 

0     +1     4-2     +3     +4     .... 


30         LANGUAGE   OF   ALGEBRA.      DEFINITIONS. 

while  the  corresponding  algebraic  series  is 

...     _4     -3     -2     -1     0     +1     +2     +3    +4    ... 

In  comparing  the  terms  of  this  double  algebraic  series, 
we  adopt  the  convention  that  they  are  here  arranged  in 
ths  order  of  magnitude,  so  that 

-3<-2<-l<0<l<2<3, 

and  in  general,  a  and  h  being  positive  numbers, 

—  a  <  —  5.  if  a  >  &.  [See  Art.  47.] 

33.  ^The  magnitude  of  a  quantity  considered  inde- 
pendently of  its  quality,  or  of  its  sign,  is  called  its 
absolute  magnitude,  or  its  absolute  value. 

Thus  a  rise  of  4  feet  and  a  fall  of  4  feet  are  equal  in  absolute 
magnitude  ;  so  also  +  4  and  ~  4  are  equal  in  absolute  magnitude, 
whatever  the  unit  may  be. 

34.  Note.  —  Although  there  are  many  signs  used  in  algebra, 
the  name  sign  is  often  used  to  denote  the  two  signs  +  and  —  ex- 
clusively. When  the  sign  of  a  quantity  is  spoken  of,  it  means  the 
+  or  —  sign  which  is  prefixed  to  it ;  and  when  we  are  directed  to 
change  the  signs  of  an  expression,  it  means  that  we  are  to  change 
the  -|-  or  —  before  every  term  into  —  and  -l- ,  respectively. 

EXAMPLES    VI. 
Calculate  the  values  of 

1.    5  _  3  _  4.  2.     5  X  3  -  19.  3.   21-^  3  -  9. 

If  a  =  1,  &  =  2,  and  c  =  3,  find  the  values  of 

4.  a  -  2  6.  6.    6c  -  11  a.  8.    6  -4-  a  -  c. 

5.  ab  —  c.  t.    ab  -be.  9,    a  ^  b  —  c. 


LANGUAGE   OF   ALGEBRA.      DEFINITIONS.         31 

10.  The  reading  of  a  thermometer  was  +  40°,  and  its  mercury 
column  then  fell  53°.     What  was  its  reading  after  the  change  ? 

11.  If  «  =  2,  5  =  3,  and  c  =  —  10,  which  is  the  greater,  axh 
or  c  ? 

12.  Can  the  result  of  subtracting  one  positive  number  from 
another  ever  be  greater  than  the  minuend  ? 


32  ADDITION.      SUBTRACTION.      BRACKETS. 


CHAPTER  III. 

Addition.    Subtraction.     Brackets, 
addition. 

35.  The  process  of  finding  the  result  when  two  or 
more  quantities  are  taken  together  is  called  addition,  and 
the  result  is  called  the  sum. 

Since  a  positive  quantity  produces  an  increase,  and  a 
negative  quantity  produces  a  decrease,  to  add  a  positive 
quantity  we  must  add  its  absolute  magnitude,  and  to  add 
a  negative  quantity  we  must  subtract  its  absolute  mag- 
nitude. 

Thus,  when  we  add  +  4  to  +  6,  we  get  +  6  -f-  4  =  + 10 ; 
and  when  we  add  —  4  to  -|-  10,  we  get  +  10  —  4  =  +  6. 

So  also,  when  we  add  4-  6  to  a,  we  get  a  +  6 ;  and  when 
we  add  —  6  to  a,  we  get  a  —  b. 

Hence  a  +  (+6)  =  a  +  6  and  a-f  (— 6)  =  a  —  b. 

We  therefore  have  the  following  rule  for  the  addition 
of  any  term :  To  add  any  term,  affix  it  to  the  expression 
to  which  it  is  to  be  added,  with  its  sign  unchanged. 

Ex.  A  "boy  played  two  games ;  in  the  first  game  he  won  6 
points,  and  in  the  second  he  won  —  4  points  (that  is  he  lost  4). 
How  many  did  he  win  altogether  ? 

The  total  gain  in  the  two  games  together  is  what  is  meant  in 
algebra  by  the  sum  of  the  gains. 

To  obtain  the  total  gain,  we  must  add  —  4  to  6,  and  this  opera- 
tion is  indicated  by  6  +  (—  4),  which  by  the  above  is  6  —  4  =  2. 


ADDITION.      SUBTRACTION.      BRACKETS.  3d 

When  numerical  values  are  given  to  a  and  to  b,  the 
numerical  values  of  a  +  6  and  a  —  b  can  be  found ;  but 
until  we  know  what  numbers  a  and  b  represent,  we  can- 
not take  any  further  step,  and  the  process  is  considered 
to  be  algebraically  complete. 

36.  It  should  be  noticed  that  when  b  is  greater  than 
a,  the  arithmetical  operation  denoted  by  a  —  5  is  im- 
possible ;  for  we  cannot  take  any  number  from  a  smaller 
number. 

Thus,  if  a  =  3  and  b  =  5,  a  —  b  will  be  3  —  5,  and  we 
cannot  take  5  from  3.  But  to  subtract  5  is  the  same  as 
to  subtract  3  and  2  in  succession,  so  that  3  —  5  =  3  —  3 
—  2  =  —  2 :  we  then  consider  that  —  2  is  2  which  is  to  be 
subtracted  from  some  other  algebraical  expression,  or 
that  —  2  is  two  units  of  the  kind  opposite  to  that  repre- 
sented by  2.  And  if  —  2  is  a  Jinal  result,  the  latter  is 
the  only  view  that  can  be  taken. 

In  some  particular  cases  the  quantities  may  be  such 
that  a  negative  result  is  without  meaning ;  for  instance, 
if  we  have  to  find  the  population  of  a  town  from  certain 
given  conditions  ;  in  this  case  the  occurrence  of  a  nega- 
tive result  would  show  that  the  given  conditions  could 
not  be  satisfied,  but  so  also  in  this  case  would  the  occur- 
rence of  a  fractional  result. 

EXAMPLES  VII. 
Find  the  sum  of 

1.  4  and  -  3.        6.   3  and  -  11.  9.   2  a  and  -  3  6. 

2.  3  and  -  2.        6.-3  and  -  9.  10.  -  3  a  and-2  b. 

3.  6and-3.        7.    6,  -  2,  and  7.  11.    6a, -6  6,  and -2c. 

4.  7  and -8.        8.    -  3,  -  2,  and  5.        12.   -3  a, -4  6,  and  7  c. 

c 


34  ADDITION.      SUBTRACTION.      BRACKETS. 

37.  It  follows  at  once  from  the  nature  of  addition 
that  ^,the  sum  of  two  or  more  algebraical  quantities, 
whether  positive  or  negative,  is  the  same  in  tohatever 
order  the  quantities  may  he  taken.  This  is  known  as  the 
Law  of  Commutation  in  addition,  j 

For  example,  to  find  how  much  a  man  is  worth,  we  can  take 
into  account  the  different  items  of  property  ;der)is  "being  con- 
sidered as  negative)  in  any  order  we  please. 

/  It  also  follows  that  to  add  any  algebraical  expression 
as  a  whole  gives  the  same  result  as  to  aCfd  its  terms  sepa- 
rately. This  is  known  as  the  Law  of  Association  in 
addition,  But  to  add  any  term,  we  have  only  to  write 
it  down,  with  its  sign  unchanged,  after  the  expression 
to  which  it  is  to  be  added. 

We  have  therefore  the  following 

Eule.  To  add  two  or  more  algebraical  expressions, 
write  down  all  the  terms  in  succession  ivith  their  signs 
unchanged. 

For  example,  the  sum  of  a  +  &  and  c  —  d  is  a  +  b-\-c  —  d;  also 
the  sum  of  a  —  6  +  c  and  — d  +  e  — fisa  —  b  +  c  —  d-^e—f. 

38.  If  some  of  the  terms  which  are  to  be  added  are 
^like'  terms,  we  can  and  must  collect  all  such  terms 
together  before  the  process  of  addition  is  considered  to 
be  complete.     Of  this  we  have  three  cases,  as  follows : 

I.  The  sum  of  '  like '  terms  which  have  the  same  sign 
is  a  'like'  term  which  has  the  same  sign,  and  whose 
coefficient  is  the  arithmetical  sum  of  their  numerical 
coefficients. 

For  example,  to  add  2  a  and  5  a  in  succession  gives  the  same 
result,  whatever  a  may  be,  as  to  add  7  a ;  that  is,  +  2  a  +  5  a  =  +  7  a. 


ADDITION.      SUBTRACTION.      BRACKETS.  35 

Also  to  subtract  2  ab  and  5  ab  in  succession  gives  the  same  result 
as  to  subtract  7  ab  ;  that  is,  —2ab  —  ^ab=—7  ab. 

II.  The  sum  of  two  4ike'  terms  whose  signs  ara 
different  is  a  *  like  ^  term  whose  coefficient  is  the  arith- 
metical difference  of  their  numerical  coefficients  and 
whose  sign  is  that  of  the  greater. 

For  example,  +6a  —  3a=+2a  +  3a  —  3a=+2a; 
also  -\-3ab  -  dab=-\-3ab-Sab  -2ab  =  -2ab. 

III.  If  there  are  several  'like'  terms  some  of  which 
are  positive  and  some  negative,  the  positive  terms  can  be 
collected  into  one  sum  by  I.,  as  also  the  negative  terms : 
the  final  sum  is  then  obtained  by  II. 

Thus  any  number  of  'like'  terms  can  be  reduced  to 
one  term. 

Ex.  1.   Add  2  a  +  3 6  and  a -5b. 

The  sum  is2a  +  36  +  a  —  56;  or,  since  the  terras  can  be  taken 
in  any  order,  the  sum  is  2a  +  a  +  36  —  56  =  3a  —  2  6,  by  col- 
lecting the  like  terms. 

Ex.  2.  Find  the  sum  of  6  a2  _  6  a6  +  462,  2  62  -  a6  -  a^,  and 
6  a6  -  9  62  -  4  a^. 

The  sum  is 
6a2  -  6  a6  4-  4  62  +  2  62  -  a6  -  a2  4  6a6  -  9  62  -  4a2 

=  6  a2  -  a2  -  4  a2  -  6  a6  -  a6  +  5  a6  +  4  62  +  2  62  -  9  6-2. 

The  terms  6  a%  —  a^^  and  —  4  a2  can  be  combined  mentally ; 
and  we  have  a"^.     Similarly  we  have  —  2  a6  and  —  3  62. 
Thus  the  required  sum  is  a2  —  2  a6  —  3  62. 

39.  It  is*  best  for  beginners  to  place  the  different  sets 
of  'like'  terms  in  vertical  columns;  so  that  the  last 
example  would  be  put  down  in  the  following  form,  the 


36  ADDITION.      SUBTRACTION.      BRACKETS. 

-f  sign  being  put  before  the  terms  26^  and  5ab,  which 
have  no  signs :  and  then  the  sets  of  '  like '  terms  can  be 
combined  mentally. 

6a2_6a&  +  462 
-  a"^  -  a&  +  2  62 
-4a2^_5a5_952 


^2  -  2  a?)  -  3  62 


EXAMPLES  VIII. 


Simplify  the  following  by  combining  '  like '  terms : 

1.  2a+6  +  3c  +  26  +  3a  +  2c  +  5a  +  c. 

2.  6a-36  +  2c-4a-3c  +  26. 

3.  5a2_3a4-6-4a2  +  6a  +  3. 

4.  7  a3  _  4  a  +  9  -  3  a2  +  2  a  +  7  -  3  a3  _  16. 

5.  5 a3  -  4 a26  +  3a52  _  5 «3  _  4 ^25  _  3 53. 

6.  3a2  +  6a6-462-2a2_4a6  +  362-a2-2a6  +  62. 
Add  together 

7.  a  +  6  and  a  —  b.  9.   ^a  +  ^6  and  -  ^ a  +  ^  6> 

8.  2x  —  y  and  2  x  +  y.  10.   ^  a  +  |  6  and  f  6  -  J  a. 

11.  a^  —  a  and  a^  _}_  q;. 

12.  a  +  a2  +  4a3and2a3_a2-4a. 

13.  m2  +  mn  +  n^  and  w2  —  mn  —  w2. 

14.  3p2  _|.  5^g  _  6  g2  and  5  ^2  _  4^^  _  3^2. 

15.  3  a2  -  2  a6  +  6'2  and  a2  _  2  ^6  -  |  6^. 

16.  2  a  +  6  -  3  c,  2  6  +  c  -  3  «,  and  2  c  +  a  -  3  6. 

17.  4  a  -  3  6  -  c,  4  6  —  3  c  -  a,  and  4  c  -  3  a  —  6. 

18.  4  a2  -  3  a6  +  62,  4  a6  -  3  62  +  a^,  and  4  62  -  3  a2  4.  ab. 

19.  a  —  ^6  +  ^c,  6-^c  +  ^«,  and  c  -  J  a  +  ^  6. 

20.  4X-22/  +  1,  -3x  +  2-y,  anda;  +  32/-3. 

21.  -4a- 6  +  2,  2  +  8a- 56,  and -a -46- 2. 


^  Q^  --^DITION.      SUBTRACTION.      BRACKETS.  37 

1  /     \       ■  ■     1   . 

22.  y?^1  x'^y  -  2 x«/2,  x'^y  -  3 xy"^  -  y^,  and  Sxy^-2y^-  x^. 

23.  a3  +  4&3_5c3  +  3a&c,  b^  + '^c^  -  3a^  +  Qabc,  and 

c3  +  4  cjs  _  5  53  _  9  Qjftc. 

24.  5  a3  _  2  a25  4.  9  a&2  +  17  &3^  _  2  a^  +  5  a^ft  -  4  a62  -  12  63, 

&3  _  4  a62  -  5  a26  -  a3,  and  2  a^ft  _  2  a3  -  6  63  _  a62. 

25.  I  a^  -  f  a25  +  3  53^  ^^s  _  2  ^52  _  3  ^,3^  and  i  a%  -lah^-\  68. 

26.  3x3-4x  +  5,  2x2 -6x+  7,  6x3-2x2 -2x,  and 

3  +  8x-4x3. 

27.  x3  -  3ax2  +  5  a2x  -  a3, 2x3  +  4ax2  -  6a2x,  6ax2  _  3a2x+  a*, 

and  -  2  x3  +  4  a2x  -  5  a3. 

28.  3x2  +  2/2  -  3y2  -  02,  2xy  -  Zy"^  +  ^yz,  and 

-  4  x2  -  2  xy  +  y2  ^  ^2. 

.    29.   Show    that,    if     x  =  6  +  2c-3a,    y  =  c  +  2a-3  6,    and 
s  =  a  +  26  —  3c;  then  will  x-\-  y  -{■  z  =  0. 

30.  Show  that,  if  a  =  5x  —  3y  —  2^?,  6  =  5?/  —  3^  —  2 x,  and 
c  =  bz  —  Zx  —  2y',  then  will  a  +  6  +  c  =  0. 

SUBTRACTION. 

40.  Since  subtraction  is  the  inverse  of  addition  [Art.  9], 
an  addition  and  a  subtraction  of  the  same  quantity  pro- 
duces no  effect.  Hence  a  is  the  same  as  a-f  6  — 6.  Now 
if  we  take  +  h  from  a  +  h  —  h^  what  is  left  is  a  —  h.  So 
that  if  we  take  away  +  h  from  a,  the  remainder  is  a  —  6 ; 
that  is 

a  —  (+6)  =  a  —  6. 

Again,  if  we  take  away  —  h  from  a  +  &  —  6,  what  is 
left  is  a  4-  6.  So  that  if  we  take  away  —  h  from  a,  the 
remainder  is  a  +  6 ;  that  is, 

a  —  (  —  6)  =  a  +  6. 


38  ADDITION.      SUBTRACTION.      BRACKETS. 

Thus  to  give  a  numerical  example, 

+  10-(+4)  =  +  10-4  =  6, 
and  +    6-(-4)  =  +   6  +  4  =  10. 

We  therefore  have  the  following  rule  for  the  sub- 
traction of  any  term :  to  subtract  any  term,  affix  it  to  the 
expression  from  which  it  is  to  be  subtracted,  but  with  its 
sign  changed. 


EXAMPLES  IX. 

Subtract 

1.  3  from  —  4. 

3.-6  from  4.              5.-6  from  a. 

2.-4  from  3. 

4.  a  from  —  6.              6.   —  a  from  —  6, 

Show  that : 

7.   -4-(+3)  =  - 

-7.                 10.   -b-(+a)  =  -b-a. 

8.  3 -(-4)=  7. 

11.  -b-i-a)  =  -b  +  a. 

9.  4 -(-6)=  10. 

12.  _(_6)_(_a)=6  +  a. 

41.  Since  subtraction  au^  addition  are  inverse  opera- 
tions, and  since  we  know  that  to  add  any  algebraical 
expression  as  a  whole  gives  the  same  result  as  to  add  its 
terms  separately,  it  follows  that  to  subtract  an  alge- 
braical expression  as  a  whole  is  the  same  as  to  subtract 
the  terms  in  succession.     (Law  of  Association.) 

We  have  therefore  the  following 

Kule.  To  subtract  any  algebraical  expression  from  any 
other,  write  down  its  terms  in  succession  with  their  signs 
changed,  after  that  other. 

Thus,   if2a-f-&  —  4c    be  subtracted  from    3  a  —  4  6  -f  c,  the 
result  is 
3a-4&-fc-2a-&+4c=3a-2a-46-6  +  c-f4c=a-5  6  +  5c. 


ADDITION.      SUBTRACTION.      BRACKETS.  6\) 

The  law  of  commutation  for  subtraction  is  the  same 
as  that  for  addition,  and  has  been  enunciated  in  Art.  37 ; 
but  in  applying  it,  the  signs,  as  well  as  the  letters  to 
which  they  are  prefixed,  must  be  interchanged.     Thus, 

+  a  — 6  =  —l)-^a 
is  an  immediate  consequence  of 

+  a  +  (-&)  =  +  (-&)  +  a. 

42.  The  expression  which  is  to  be  subtracted  is  often 
placed  under  that  from  which  it  is  to  be  taken,  *  like ' 
terms  being  for  convenience  placed  under  one  another . 
and  the  signs  of  the  lower  line  are  changed  mentally 
before  combining  '  like '  terms. 

Thus  the  example  considered  in  Art.  41  would  be  written  as 
follows : 

3a-46+     c 
2a+     6-4c 

a  —  66  +  6c 

The  terms  of  the  result  being  obtained  by  combining  mentally 
3  a  and  —2  a,  —  4  6  and  —  6,  and  c  and  -1-  4  c. 
As  another  example,  if  we  have  to  subtract 

Sa^-4a%-\-2 ab'^  -  b^  from  4a3  +  a^b  -  ab\ 

the  process  is  written  as  follows : 

4  a8  +    a^b  -    ab"^ 
3a8-4a26  +  2a62_68 

a*  +  6  a26  -  3  ab'^  +  b^ 

The  terms  of  the  result  being  obtained  by  combining  4  a^  and 
3a3,  a^b  and  +  ^ffib,  -  ab^  and  -  2a6^  0  and  +  b^ 

43.  We  have  hitherto  supposed  that  the  letters  used 
to  represent  quantities  were  restricted  to  positive  values  ; 


40  ADDITION.      SUBTRACTION.      BRACKETS. 

it  would,  however,  be  very  inconvenient  to  retain  this 
restriction.  In  what  follows  therefore  it  must  always  be 
understood,  unless  the  contrary  is  expressly  stated,  that 
each  letter  may  have  any  value  positive  or  negative. 

Since  any  letter  may  stand  for  either  a  positive  or  a 
negative  quantity,  a  term  preceded  by  the  sign  +  is  not 
necessarily  a  positive  quantity  in  reality ;  such  terms 
are,  however,  still  to  be  called  positive  terms,  because  they 
are  so  in  appearance  ;  and  the  terms  preceded  by  the  sign 
—  are  similarly  called  negative  terms. 

44.  We  must  now  carefully  examine  whether  terms 
can  be  added  and  subtracted  without  knowing  whether  the 
letters  really  represent  positive  or  negative  quantities. 

Now  we  have  seen  in  Arts.  35  and  40,  that  when  h  is 
really  positive, 

4-(+5)  =  +  6,   +(_6)  =  _6,   _(4-6)  =  _6, 

and  —  (  —  6)  =  +  6  ; 

and  we  have  to  see  whether  the  same  laws  hold  good 
although  h  may  really  be  negative. 

If  h  be  really  negative  and  equal  to  —  c,  where  c  is  posi- 
tive, then  4-5  =  +  (  —  c)  =  —  c,  and  — 6  =  —  (—  c)  =  +  c, 
since  c  is  positive.  Hence,  putting  —  c  for  +  h,  and  -|-  c 
for  —  6,  the  laws  expressed  above  will  be  true,  although 
b  is  negative,  provided 

+  (-c)  =  -c,  4-(4-c)  =  +  c,  _(_c)=  +  c, 

and  —  ( +  c)  =  —  c, 

are  true  for  all  positive  values  of  c,  and  this  we  know  is 
the  case. 


ADDITION.      SUBTRACTION.      BRACKETS.  41 

Hence  terms  are  added  or  subtracted  in  precisely  the 
same  way  whether  the  letters  really  stand  for  positive  or  for 
negative  quantities. 

45.  Zero,  whose  symbol  is  0,  may  now  be  defined  as  a 
difference,  in  the  form 

a  —  a  =  0. 

When  it  appears  in  any  algebraic  expression,  as  a  term 
in  the  form  0,  or  as  a  pair  of  terms  in  the  form  a  — a, 
or  as  a  series  of  terms  reducible  to  either  of  these  forms, 
it  may  obviously  be  erased  without  in  any  way  affecting 
the  numerical  value  of  the  whole  expression. 

46.  Def.  The  algebraical  difference  between  any  two 
quantities  a  and  h  is  the  result  obtained  by  subtracting 
the  second  from  the^rs^ 

For  example,  tlie  algebraical  difference  of  5  and  4  is  5  —  4  =  1, 
and  the  algebraical  difference  of  4  and  5is4  —  5  =  — 1. 

The  algebraical  difference  between  tvo  quantities  de- 
pends upon  the  order  in  which  they  are  given,  and  may 
therefore  not  be  the  same  as  the  arithmetical  difference, 
which  is  the  result  obtained  by  subtracting  the  less  from 
the  greater. 

The  symbol  a  ~  6  is  used  to  denote  the  arithmetical 
difference  of  a  and  h. 

47.  Def.  One  quantity  a  is  said  to  be  greater  than 
another  quantity  6,'  when  the  algebraical  difference,  a  —  b, 
is  positive. 

From  this  definition  it  is  easy  to  see  that  in  the  series  1,  2,  3, 
4,  etc.,  each  number  is  greater  than  the  one  before  it ;  and  that, 


42  ADDITION.      SUBTRACTION.      BRACKETS. 

in  the  series  —  1,  —  2,  —  3,  —  4,  etc.,  each  number  is  less  than  the 
one  before  it.  Thus  7,  6,  1,  0,  —  5,  -  7  are  in  descending  order  of 
magnitude.     [See  Art.  32.] 

EXAMPLES  X. 
Subtract 

1.  a  —  b  from  a  +  6.  ^.    ^x  —  ly  from  ^x  —  ^y. 

2.  2a -3&  from  3a -26.  6.  3x  -  4x2  from  4x  -  3x2. 

3.  2 X  +  ?/  from  2x-y.  6.   x^  -  2  from  1  -  2 x". 

7.  4a-26  +  3c  from  2c+4&-3a. 

8.  4  a2  -  2 a6  +  3 &2  from  2b'^  +  4.ab  -  Sa^ 

9.  3  x2  -  4  X  +  2  from  x2  +  6  x  -  7. 

10.  3x3-2x2  + 5  from  3x3-2x4- 5. 

11.  a  —  ^b  —  ^c  from  b  —  ^c  —  ^a. 

12.  ^x2  -  ixy  +  |?/2  from  1 2/2- ix?/  +  ^x2. 

Find  the  difference  between 

13.  a +  26  and  a -26.  14.   3a  -  76  and  7a  -  36. 

15.  a2  +  a6  +  62  and  a2  -  a6  +  62. 

16.  x2  -  3  xy  and  3  x2  -  4  xy. 

17.  3x3  +  5x2?/  +  4 jcy2  and  4x^y  +  6 xy^  +  7 y^ 

18.  2  x8  -  7  x2?/  +  9  x?/2  -  4  ?/3  and  4  x^  -  xSy  +  9  x?/2  -  4 1/«. 

19.  What  must  be  added  to  2  a  —  3  6  in  order  that  the  sum  may 
be  4  a  -  6  6  ? 

20.  What  must  be  added  to  a2  +  62  in  order  that  the  sum  may 
be  2  a2  -  a6  ? 

21.  What  must  be  added  to  5  a6  —  2  6c  +  3  ca  in  order  that  the 
sum  may  be  7  a6  +  2  ca  ? 

22.  What  must  be  added  to  a2  +  3  62  +  2  c2  that  the  sum  may 
be62-3a2? 


ADDITION.      SUBTRACTION.      BRACKETS.  43 

23.  Subtract  from  3  a  -  4  6  the  sum  of2a  +  7  6,  -4a-6&, 
and  6  a  —  5b. 

24.  Subtract  from  3  x^  -  2  a;  +  7  the  sum  of  x^  -  x  +  9,  2x^ 
-f  7  X  -  6,  and  3  a;2  -  4  X  -  5. 

26.  Subtract  the  sum  of  a'^-4:ah-^  b\  ab-ib'^-Sa^,  and 
62  _  4  ^2  _  3  a6  from  2  a^ -^  2  b^  +  2  ab. 

BRACKETS. 

48.  To  indicate  that  any  algebraical  expression  is  to 
be  added  as  a  whole,  it  is  put  between  brackets  (paren- 
theses, braces)  with  the  +  sign  prefixed.  But,  as  we 
have  seen,  to  add  any  expression,  we  have  only  to  write 
down  the  terms  in  succession,  with  their  signs  unchanged. 

Hence,  when  a  bracket  is  preceded  by  a  +  sign,  the 
brackets  may  be  omitted.     Thus 

4-(2a-6-fc)  =  2a-&  +  c. 

Conversely,  any  number  of  terms  in  an  expression  may 
be  enclosed  in  brackets  with  the  sign  +  placed  before 
the  bracket. 

For  example,  we  may  write 

a-26  +  c  +  2d--3e+/ 
in  the  form 

a4-(-26  +  c)  +  (2(«-3e+/), 

or  a-26  +  (c-f2d)-|-(-3e-f/). 

When  the  sign  of  the  first  term  within  the  brackets  is 
-h,  it  is  generally  omitted  for  shortness,  as  in  the  preced- 
ing example. 

49.  To  indicate  that  any  algebraical  expression  is  to 
be  subtracted  as  a  whole,  it  is  put  between  brackets,  and 


44  ADDITION.'     SUBTEACTION.      BRACKETS. 

the  —  sign  prefixed.  But,  as  we  have  seen,  to  subtract 
any  expression,  we  have  only  to  write  down  the  terms  in 
succession  with  all  their  signs  changed. 

Hence,  when  a  bracket  is  preceded  by  the  —  sign,  the 
brackets  may  be  omitted,  provided  that  the  signs  of  all 
the  terms  within  the  brackets  are  changed. 

Thus  -(2  a  —  6  +  c)  =  -2a  +  6-c. 

Conversely,  any  number  of  terms  in  an  expression  may 
be  enclosed  in  brackets  with  the  sign  —  placed  before 
the  bracket,  provided  that  we  change  the  signs  of  all  the 
terms  which  are  placed  in  the  brackets. 

Thus         a-2b-{-c  +  2d  —  3e-\'f 
may  be  written  in  the  form 

a-{2b-c)-(-2d  +  3e  -/) . 

50.  Sometimes  one  enclosure  is  put  within  another: 
in  this  case  the  different  pairs  of  brackets  must  be  of 
different  shapes  to  prevent  confusion. 

Thus  a— [6  +  Jc  — (d4-e)5]  ;  which  means  that  we 
are  to  add  to  b  the  whole  quantity  within  the  braces  \  \, 
and  then  subtract  the  result  from  a;  and  to  find  the 
quantity  within  the  braces  JJ?  ^®  must  add  d  and  e,  and 
then  subtract  the  sum  from  c. 

When  there  are  several  pairs  of  brackets,  they  may  be 
removed  one  at  a  time  by  the  rules  of  Arts.  48  and  49 ; 
and  it  is  best  for  beginners  to  remove  at  every  stage  the 
innermost  bracket. 

Thus  a  -  [6  +  |c  -  (d  +  e) }]  =  a  -  [6  +  {c  -  d  -ej] 

=  a—  [6  4-c  —  d  —  e]  =  a  —  6  —  c  +  d  +  e. 


ADDITION.      SUBTRACTION.      BRACKETS.  45 

EXAMPLES   XI. 

Simplify  the  following  expressions  by  removing  the  brackets 
and  collecting  like  terms : 

1.  (a  +  b)-(a-b).  6.  a -[a  -  {a -(2a  -  a)}]. 

2.  a-b-(a-\-b).  7.  1 -[2  -  {3 -(4  -  5)}]. 

3.  a -(&  + c)  +  (6 -c-a).     8.  a  +  6-[a-6+{a  +  6-(a-6)}]. 

4.  Sx-(y-2x)  +  (z-{-y-5x).9.  5 -[4  +  {5 -(4  +  S^^l)}]. 


6.  x-{y  -(z  -  x)}.  10.   x-[y-{z-{x-y  -  «)}]. 


11.   3x-{2y  +  50  -  3x  +  y}. 


12.   {2x-(i^y-Sz  -h  7)]  -li  +  {X  -(Sy  -\-2z  +  5)]]. 


13.    [2 a -{36+ (4c --36 +  2 a)}]. 


14.  x-(y  -  z)-\-{2z-  Sy-6x]. 

15.  a  -  2  6  -  {3  a  -  (6  -  c)  -  5  c}. 

16.  a-[36  +  {3c-((f-6)+a}-2a]. 

17.  3a -[26  — {4c -12a- (46  -  8c)}  -  (6  6  -  12c)]. 

18.  {2x-(Sy-7z)  +  (Sx-2y  +  9z)} 

-  {(y  -  50)-(3x  ~y-2z)  +  Sz], 

19.  a2-(3a6-462)-(2a2-3a6  +  662) 


-{5  62-(3a6-7a2-62)}. 
20.   (m2  -  w2)  -  {3  WW  -  (5  n2  -  m^)} 

+  [n2  -  {3  mw  -  (5  m^  -  6  n2)}  +  8  mn]. 


46  MULTIPLICATION. 


CHAPTER  IV. 

Multiplication. 

51.  In  arithmetic,  multiplication  is  first  defined  to  be 
the  taking  one  number  as  many  times  as  there  are  units 
in  another.  Thus,  to  multiply  5  by  4,  we  take  as  many 
fives  as  there  are  units  in  four.  As  soon,  however,  as 
fractional  numbers  are  considered,  it  is  found  necessary 
to  modify  somewhat  the  meaning  of  multiplication,  for 
by  the  original  definition  we  can  only  multiply  by  whole 
numbers.  The  following  is  therefore  taken  as  the  defi- 
nition of  multiplication : 

Def.  To  multiply  one  number  by  a  second  we  do  to  the 
first  what  is  done  to  unity  to  obtain  the  second. 

Thus  4isl-f-l  +  l  +  l; 

.-.  5  X  4  is  5  -f  5  +  5  +  5. 

Again,  to  multiply  4  by  f,  we  must  do  to  -f-.  what  is 

done  to  unity  to  obtain  J;  that  is,  we  must  divide  ^  into 

four  equal  parts  and  take  three  of  those  parts.     Each  of 

5 
the  parts  into  which  4  is  to  be  divided  will  be -,  and 

5x3  '^X^ 

by  taking  three  of  these  we  get  - — -• 

So  also  (-5)x4=-5+(-5)-f-(-5)-i-(-5) 
=  _5_5_5-5 
=  -20. 


MULTIPLICATION.  47 

With  the  above  definition  multiplication  by  a  negative 
quantity  presents  no  difficulty. 

For  example,  to  multiply  4  by  —  5.  Since  to^subtract 
5  by  one  subtraction  is  the  same  as  to  subtract  five  units 
successively, 

...  4x  (-5)  = -4 -4 -4- 4- 4 
=  -20. 
Again,  to  multiply  —  5  by  —  4.     Since 
-4  =  -l-l-l-l; 
.-.  (-5)x(-4)  =  -(-5)-(-5)-(-5)-(-5) 

=  +  54-5  +  5  +  5  [Art.  40.] 

=  +  20. 

We  can  proceed  in  a  similar  manner  for  any  other 
numbers,  whether  integral  or  fractional,  positive  or  nega- 
tive.    Hence  we  have  the  following  laws : 

ax        6  =  +  a6  .  .  .  (i.) 

(—a)  X        6  =  — a6  .  .  .  (ii.) 

ax{—b)  =  —ah  .  .  .  (iii.) 

(-a)x{-b)  =  -\-ab  .  .  .  (iv.) 

The  rule  by  which  we  determine  the  signs  of  the  prod- 
ucts is  called  the  Law  of  Signs;  this  law  is  sometimes 
enunciated  briefly  as  follows :  like  signs  give  +,  and 
unlike  signs  —  . 

52.   The  factors  of  a  product  may  be  taken  in  any  order. 

It  is  proved  in  arithmetic  that  when  one  number, 
whether  integral  or  fractional,  is  multiplied  by  a  second. 


48  MULTIPLICATION. 

the  result  is  the  same  as  when  the  second  is  multiplied 
by  the  first. 

The  proof  is  as  follows : 

First,  when  the  numbers  are  integers,  a  and  h  suppose, 
write  down  a  series  of  rows  of  dots,  putting  a  dots  in  each 
row,  and  take  h  rows,  writing  the  dots  under  one  another 
as  in  the  following  arrangement : 


a  m  a  row 


h  rows. 

Then,  counting  by  rows,  the  whole  number  of  the  dots 
is  a  repeated  h  times ;  that  is,  a  xb.  Also,  counting  by 
columns,  the  whole  number  of  the  dots  is  b  repeated  a 
times ;  that  is,  b  x  a.     Hence,  when  a  and  b  are  integer s, 

a  xb  =  b  xa. 

Next,  when  the  numbers  are  fractional,  for  example, 

5x3 
j-  and  J,  we  prove  as  in  Art.  51  that  ^  x  f  =  = — j-    And, 

5x3     3x5 

by  the  above  proof  for  integers,  =  ;  hence 

7x44x7 

T  X  ^  —  4    X  y. 

Hence  ab  =  ba  for  all  positive  values  of  a  and  b ;  and 
being  true  for  any  positive  values  of  a  and  b,  it  must  be 
true  for  all  values,  whether  positive  or  negative;  for, 
from  the  preceding  article,  the   absolute  value  of  the 


MULTIPLICATION.  49 

product  is  independent  of  the  signs,  and  the  sign  of  the 
product  is  independent  of  the  order  of  the  factors. 
Hence  for  all  values  of  a  and  h  we  have 

ab  =  ba  .     .     .     .     .     .     .     (i.) 

This  is  the  Law  of  Oommutation  in  multiplication. 

If  in  the  above  arrangement  of  dots  we  put  c  in  the 
place  of  each  of  the  dots,  the  whole  number  of  c's  will  be 
ab ;  also  the  number  of  c's  in  the  first  row  will  be  a,  and 
this  is  repeated  b  times.  Hence,  when  a  and  b  are 
integers,  c  repeated  ab  times  gives  the  same  result  as  c 
repeated  a  times  and  this  repeated  b  times.  So  that  to 
multiply  by  any  two  whole  numbers  in  succession  gives 
the  same  result  as  to  multiply  at  once  by  their  product, 
and  the  proposition  can,  as  before,  be  then  proved  to  be 
true  without  restriction  to  whole  numbers  or  to  positive 
values. 

Thus,  for  all  values  of  a,  6,  and  c,  we  have 

a  xb  X  c  =  ax  (be) (ii.) 

This  is  the  Law  of  Association  in  multiplication. 

From  (i.)  and  (ii.)  it  follows  that  the  factors  of  a 
product  may  be  taken  in  any  order  without  altering  the 
result,  however  many  factors  there  may  be. 

63.  Since  the  factors  of  a  product  may  be  taken  in 
any  order,  we  are  able  to  simplify  many  products. 

For  example : 

3a  x4a  =  3x4xaxa  =  12a2, 
(-3a)x(-46)  =  4-  3a  X  46=+3x  4  X  a  X  &  =  12a6, 
(a6)2  =:ahxah  =  axaxhy.h  =  a^b^, 
(^  a)2  =  V2  a  X  V2  «  =  V2  X  V2  X  a  X  a  =  2  a2. 


50  MULTIPLICATION. 

Although  the  order  of  the  factors  in  a  product  is 
indifferent,  a  factor  expressed  in  figures  is  always  put 
first,  and  the  letters  are  usually  arranged  in  alphabetical 
order. 

54.  By  definition,  c?  —  aa,  a^  =  aaa,  a^  =  aaaa,  etc. 
Hence  a^  X  a^  =  aa  x  aaa  =  a^  =  a^+^ ;     , 
also            .          a^  X  a^  =  aaa  x  aaaaa  =  a^  =  a^+* ; 
and  a  x  a'*  =  a  x  aaaa  —  a^  =  a^+'*. 

In  the  above  examples  we  see  that  the  index  of  the 
product  of  two  powers  of  the  same  letter  is  equal  to  the 
sum  of  the  indices  of  the  factors.  We  can  prove  in  the 
following  manner  that  the  above  is  true  whenever  the 
indices  are  positive  integers : 

Since,  by  definition, 

a"*  =  aaaa  •  •  •  to  m  factors, 
and  a"  =  aaaa  •  •  •  to  n  factors  ; 

.-.  a*"  X  a"  =  (aaa  •  •  •  to  m  factors)  x  (aaa  •  •  •  to  n  factors) 
=  aaa   •  •  •  to  m  +  w  factors 
=  a'"+'*,  by  definition. 

Thus,  for  any  positive  values  of  m  and  n 

a"^  X  a''  =  a"'+^ 

This  result  is  called  the  Index  Law. 

55.  Product  of  Monomial  Expressions.  The  results  arrived 
at  in  the  preceding  articles  will  enable  us  to  find  the 
product  of  any  monomial  expressions.  These  results 
are : 

(i.)  The   sign   of  the   product   of  two   quantities   is 


MULTIPLICATION.  51 

4-  when  the  factors  are  both  positive  or  both  negative ; 
and  the  sign  of  the  product  is  —  when  one  factor  is 
positive  and  the  other  negative. 

(ii.)  The  factors  of  a  product  may  be  taken  in  any 
order. 

(iii.)  The  index  of  the  product  of  any  two  powers 
of  the  same  quantity  is  the  sum  of  the  indices  of  the 
factors. 

Ex.  1.  Multiply  3  a262  by  6  a^fts. 

3 a262  x6 ^253  =  3  X  6  X  a2  X  a2  X ?)2  X  68,       from  (ii.) 
=  18  a2  +  252  +  3=18  a*65,  from  (iii.) 

Ex.  2.   Multiply  -  3  a26  by  -  6  ah^. 

(-3 a2&)  X  (-  5 a&5)  =  +  3 a26  x  5 ah^,  from  (i.) 

=  3x5xa2xax6x6®,  from  (ii.) 

=  16  a2  +  ifti  +  6  =  16  a^h^,  from  (iii.) 

Ex.  3.   Multiply  2  a^h^d^  by  -  6  a^h. 

The  whole  work  of  finding  the  product  of  two  monomial  expres- 
sions can  be  performed  mentally  and  the  result  written  down  at 
once.  First  put  down  the  sign  of  the  result;  then  multiply  the 
numerical  coefficients  to  find  the  numerical  coefficient  of  the 
product ;  then  take  each  letter  whiclf  occurs  to  a  power  whose 
index  is  the  sum  of  the  indices  of  that  letter  in  the  factors.  Thus 
the  product  of 

2  a'^h^c^  and  -  6  a^h  is  -  10  a'h^c^. 

Ex.  4.   Find  the  cube  of  2  a^h. 

The  cube  is  2  a2?,  x  2  a26  x  2  a^h  =  8  a^b\ 


52  MULTIPLICATION. 

EXAMPLES  XII. 
Multiply 

1.  3  a  by  6  a.  10.  2  a  by  -4  &. 

2.  5  a2  by  7  a.  11.  3  &  by  -  4  a. 

3.  2  a^  by  5  a^.  12.  a^  by  -  a. 

4.  rt6bya2&3.  13.  _6a36by4a6. 

6.  3a2&by2a62.  14.  -2ah^\>y -1  aW. 

6.  4  a63  by  7  a462.  15.  _  3  a^^c^  by  6  aftV. 

7.  3a2&c3by6a6.  16.  -  3  a&2c  by  2  a&3c2. 

8.  5  ahH^  by  3  aZ)^.  17.  _  2  a;?/*  by  -  5  a:V- 

9.  2  aWc  by  a6c.  18.  2  ax^y^z  by  -  5  a2icy4;23. 

19.  6  aWcH^y^z  by  -  12  a&3c4a;2?/32;5. 

20.  Find  the  values  of  (-  a)2,  (-  of,  and  (-  a)*. 

21.  Find  the  values  of  (-  ^2)2,  (-  x2)3,  and  (-  x'^y. 

22.  Find  the  values  of  (-  a6)2,  (-  ahf,  and  (-  aby. 

23.  Find  the  values  of  {a^h^Y,  {a%^y,  and  (a2&3)4. 

24.  Find  the  values  of  (-  2  a%^y,  (-  2  a354)3^  and  -(2  a^ft*)*. 

25.  Find  the  squares  of  3  aV^o^,  —  2  a3?)c4,  and  —  4  a263c5. 

26.  Find  the  cubes  of  a^^  —  «*,  ah,  and  —  a25. 

27.  Find  the  cubes  of  2  a62,  _  3  a3&2^  _  4  0,55^  and  -  7  a255c4. 

28.  Find  the  values  of  (-  a)2  x  (-  &)3,  ( -  2 a)3  x  (a2)2, 

(-  aW-Y  X  (-  a2&)3,  and  (a262)3  x  (-a63)4. 

If  a  =  2,  &  =  —  3,  c  =  —  1,  (?  ,=  0,  find  the  numerical  values  of 

29.  6a6.  35.  3  a262c3  +  5  52^4. 

30.  4a&c.  36.  2  a&2  +  3  5^2  +  4  C(Z2. 

31.  8a262c2.  37.  (a  +  6)(c+(?). 

32.  ab^U^ca.  38.  (a  -  &)2(c  -  d)2. 

33.  a3  4.  53  +  c3  +  (Z3.  39.  (a2  +  jc)  (62  +  cd;>), 

34.  a252  _|.  ^,2^2  _|.  c2(22.  40.  (a  -  6)3(c  -  a)^. 


MULTIPLICATION.  63 

56.  We  now  proceed  to  the  multiplication  of  multino- 
mial expressions. 

We  first  observe  that  any  multinomial  expression  can 
be  put  in  the  form 

a-\-b  +  c-{-  etc., 

where  a,  6,  c,  etc.,  may  be  any  quantities  positive  or  nega- 
tive. 

For  example,  the  expression  ^x^y  —  ^  xy"^  —  7  xyz,  which 
by  Art.  35  is  the  same  as  ^ary -{-{— ^xy^)-{-{—l xyz)^ 
takes  the  required  form  if  we  put  a  for  3a^y,  h  for 
—  ^xy"^,  and  o  for  —  7  xyz. 

It  therefore  follows  that  in  order  to  prove  any  theorem 
to  be  true  for  any  algebraical  expression,  it  is  only  neces- 
sary to  prove  it  for  the  expression  a  -)-  6  -f-  c  -f  etc.,  where 
a,  b,  c,  etc.,  are  supposed  to  have  any  positive  or  negative 
values. 

57.  Product  of  a  Multinomial  Expression  and  a  Monomial. 
Consider  the  product  {a  4-  b)c,  where  a,  6,  and  c  have  any 
values  whatever. 

If  c  be  a  positive  integer^  and  a  and  h  have  any  values 
whatever;  then 

{a-\-b)c=  (a-f  6)  -f-  (a-f  6)  +  {a-{-  b)  -\ repeated  c  times, 

—  a-{-b-\-a-\-b-\-a-\-b-\ 

=  a4-a+a-f  •••  repeated  c  times 

-\-b-\-b-\-b-\-  ...repeated  c  times, 

=.ac-{-bc. 

Hence,  when  c  is  a  positive  integer,  we  have 

(a-h  b)c  =  ac  -f  be. 


54  MULTIPLICATION. 

Since  division  is  the  inverse  of  multiplication,  it  fol- 
lows that  when  d  is  a  positive  integer, 

(a  +  b)-i-d  =  a-i-d-{-b-^d. 

And  since  the  operations  of  multiplication  and  division 
to  be  performed  on  a  +  6  can  be  performed  on  a  and  b 
separately,  this  must  also  be  the  case  for  the  complex 

operation  denoted  by  a  fraction  — 

Thus  (a  -{-b)c  =  ac-\-bc  for  all  positive  values  of  c; 
and  being  true  for  any  positive  value  of  c,  it  must  also 
be  true  for  any  negative  value.     For,  if 

(a  +  b)c  =  ac-{-  be, 
then  —  (a  +  b)c  =  —  ac  —  be; 

and  therefore    (a  +  6)  (—  c)  =  a(—  c)  +  ^(— c). 
Hence,  for  all  values  of  a,  b,  and  c,  we  have 
{a-]-b)c  =  ac  -\-  be. 

58.  Since  (a-{-b)c  =  ac-\-bc  for  all  values  of  a,  b,  and 
c,  it  will  be  true  when  x-\-y  is  put  in  place  of  a. 

Hence  \  {x  -\-  y)  +  blc  =  {x  -\- y)c  -\-  be  =  xc  +  ye  +  be; 
.-.   {x-\-y  +  b)c  =  xc-\-yc-\-bc. 

And  similarly 

{x-\-y-\-z-{-p-] )c  =  xc-{-yc-\-zc-\-pc-\ , 

however  many  terms  there  may  be  in  the  expression 
x-{-y-\-z-\-p-\-'". 
Thus  the  product  of  any  multinomial  expressio7i  by  a  '^ 
monomial  is  the  sum  obtained  by  multiplying  the  separate 
terms  of  the  multinomial  expression  by  the  monomial 


MULTIPLICATION.  56 

This   result  is   called  the   Distributive   Law  in   multi- 
plication. 

Ex.  1.   Multiply  a2  +  a^hy  a. 

The  result  is  a'^  x  a  +  a^  x  a  =  a^  +  a^. 

Ex.  2.    Multiply  a  —  6  by  c. 

The  result  is  a  x  c +  {~  b)x  c  =  ac  —  be, 

Ex.  3.   Multiply  -  3x2  by  a;  -  1. 

-3x2x(x-l)  =  (x-l)x(-3x2)  =  -3x8  +  3«2. 

EXAMPLES    XIII. 
Multiply 

1.  a  +  6  by  3.  8.  2  a^  -  3  a  -  4  by  -  3  a^. 

2.  2  a  -ft  by  4.  9.  2  a^  -  3  a6  +  2  62  by  a262. 

3.  3  a  -  4  &  by  6.  10.  be -\- ca -\-  ab  by  abc. 

4.  a2  +  a  by  a.  11.  2x^  -  3x2  +  5x  -  4by  -  5x2. 

5.  a2-abya8.  12.  4  -  3x2  +  3x8  -  4x*by- 6x» 

6.  a^  +  lhySa^  18.   -6a6  by  3a2-2a6  +  7  &2. 

7.  4a2_6a  +  lby  a*.  14.  -  6a86*by2a8-3a26-662. 
Simplify 

15.  2(a-6)-f  4(a  +  6).  17.   c^a -^  b)- c(a- b). 

16.  K^-2c)-|(c-2  6).       18.  7a(6-c)-26(a-c). 

19.  a262(c2  _  (f2)  +  c2d2(a2  -  6*2)  +  62c2((f2  _  a2). 

20.  2{3a6-4a(c-2  6)}. 

21.  3a -2[6- {2c -6a -2(6 -2c)} -3(6 -2c)]. 

22.  4  aS  -  [(2  63  -  3c8)  -  6  6c(6  +  c)  +  3  c(c2  +  2  62)]. 

23.  7  ac  -  2  {2  c(a- 3  6)- 3(5  c-^^6)a}. 


56  MULTIPLICATION. 

59.  Product  of  Two  Multinomial  Expressions.  We  have 
now  to  consider  the  most  general  case  of  multiplica- 
tion ;  namely,  the  multiplication  of  any  two  multinomial 
expressions. 

We  have  to  find      ' 

(a  +  &  +  c+-..)  x(x4-2/  +  2J---); 

and,  from  Art.  56,  this  includes  all  possible  cases. 

Put  Jf  for  x-^y-^z-i ;  then,  by  the   last  article, 

we  have 

(a  +  &  +  c  4-  "')M=:  aM+  hM-{-  cM+  ... 

=  Ma-\-Mb-\-  Mc  +  ...  [Art.  52.] 
=  (x-^y  +  z-{-'")a  +  {x  +  y-\-z-{-  ...)6 

+  (a;-f-2/  +  2;+...)c-f--- 

=  ax -\-  ay  -{- az -{- \- bx -{- by  +  bz -\ 

-{- ex -\- cy  +  cz -\ 1 

Hence       {a  +  b -\- c-\- "')(x -\-y -\-z-{- -") 
=  ax.-\- ay -\- az -\ \-bx-{-by-{-bz-\ \- ex -\- cy -{- cz -] 

Thus  the  product  of  any  two  multinomial  expressions  is 
the  sum  of  the  products  obtained  by  multiplying  every  term 
of  the  multiplicand  by  every  term  of  the  multiplier. 

For  example, 

(a  +  6)  (c  +  ^)  =  etc  +  6c  +  ad  -f-  bd. 
Also 
(2a+56)(3aH-26)=2ax3a  +  56x3a  +  2ax26  +  5&x26 
=  6a^-fl5a6  +  4a6  +  10  52=6a2+19a6  +  1062. 


MTJLTIPLICATION.  57 

Again,  to  find  (a  — 6)x(c  — d);  we  must  first  write 
this  in  the  form  J  a-f  (— 6)  j  Jc  +  (— d)  j,  and  we  then 
have  for  the  product 

ac  +  (  -  6)c  +  a(  -  d)  +  ( -  6)  ( -  d), 

which  by  Art.  51  is  equal  to 

ac  —  bc—  ad-{-  bd. 

Note.  In  the  rule  given  above  for  the  multiplication  of  two 
algebraical  expressions  it  must  be  borne  in  mind  that  the  terms 
include  the  prefixed  signs. 

60.    The  following  are  important  examples : 

(i.)    (a  4-  by  =  (a  +  6)  (a  -f-  6)  =  aa  -h  6a  +  a6  +  &6; 
.-.   (a  +  by  =  a'-^2ab-^b\ 

Plence,  the  square  of  the  sum  of  any  two  quantities  is 
equal  to  the  sum  of  their  squares  plus  twice  their  product. 

(ii. )   (a-by  =  (a-b)  {cl^-th)=^  +  (  -  b)  a  +  at  -  b') 
-^{-b)(-b)  =  a^-ab-ab+  b') 
.'.  (a-by=a^-2ab-{-b\ 

Hence,  the  square  of  the  difference  of  any  two  quantities 
is  equal  to  the  sum  of  their  squares  minus  twice  their 
product. 

(iii.)  (a  +  b){a-b)  =  aa  +  ba  +  a{—b)  +  b(-  b) 
=  a^  -{-  ab  —  ab  —  b^] 
.'.   {a-\-b){a-b)  =  a--b'. 

Hence,  the  product  of  the  sum  and  difference  of  any  two 
quantities  is  equal  to  the  difference  of  their  squares. 


68  MULTIPLICATIOK. 

61.   It  is  usual  to  exhibit  the  process  of  multiplication 
in  the  following  form : 

a2  +  2  a&  -  62 

a^  +  2a^h-    o?h'' 

-2a^h-4.a^h^  +  2ab^ 

o?W  +  2aW-h' 
a*  _4a262  +  4a63_54 

or  by  a  rectangular  arrangement  as  follows  : 


-2ah 

-2o?h-4.o?h^  +  2ah^ 

4-    a262  +  2a63-6* 

a*             _4a262  4.4a53_54 

The  multiplier  is  placed  under  the  multiplicand, 
arranged  horizontally,  or  to  the  left  arranged  vertically. 
The  successive  terms  of  the  multiplicand,  namely  a?, 
+  2  ah,  and  —  Ir^,  are  multiplied  by  a?,  the  first  term  on 
the  left,  or  at  the  top,  of  the  multiplier ;  and  the  products 
a^,  +  2  a^h,  and  —  a^h^,  which  are  thus  obtained,  are  put 
in  a  horizontal  row.  The  terms  of  the  multiplicand  are 
then  multiplied  by  —  2  ah,  the  second  term  of  the  multi- 
plier, and  the  products  thus  obtained  are  put  in  another 
horizontal  row,  the  terms  being  so  placed  that  ^like' 
terms  are  under  one  another.  The  terms  of  the  multi- 
plicand are  then  multiplied  by  h^,  the  last  term  of  the 
multiplier,  and  the  products  thus  obtained  are  put  in  a 
third  horizontal  row,  ^like'  terms  being  again  placed 
under  one  another.     The  final  result  is  then  obtained  by 


MULTIPLICATION. 


59 


adding  the  rows  of  partial  products ;  and  this  final  sum 
can  be  readily  written  down,  since  the  different  sets  of 
'  like '  terms  are  in  vertical  columns. 

62.   The  following  are  additional  examples  of  multipli- 
cations arranged  as  in  the  preceding  article : 

a  +  6  a  —  b  a-{-b 

a  +  6  a  —  b  a  —  b 


a^  +  ab 

a^  —  ab                  a^  +  ab 

-\-ab    +b^ 

-ab    +b^            -ab- 

-b' 

a^  +  2ab+  b^ 

a^-2ab-\-b^        a^ 

-W 

a  +^2b 

a  +      64-c 

a  -V2& 

a 

a-  -h    a6  +  a^ 

a^-]--y/2ba 

+  b 

+    ab             -{-b^  + 

be 

-^2ba-2b^ 

+  c 

-hac            + 

bc  +  c" 

2b'- 


a^'\-2ab^2ac^-b^^-2bc-\-<? 


63.  While  these  arrangements  are  helpful  to  the  be- 
ginner, the  practised  worker  in  algebra  writes  out  the 
product  of  two  simple  multinomials  without  going 
through  this  somewhat  long  and  formal  process.  Let 
the  results  in  some  of  the  examples  of  the  following 
list  be  written  out  in  this  way.  Thus,  in  order  to  write 
down  the  product  of  ic  +  1  and  ic^  -f-  a;  -|-  1,  observe  that 
there  can  be  only  four  kinds  of  terms ;  namely,  terms  in 
a^,  a;^,  x,  and  a  term  independent  of  x.  The  only  partial 
product  containing  a^  is  xy,£-\  two  partial  products, 
xy^x  and  1  x  a^,  contain  oi? ;  two,  x  x  1  and  1  x  a;,  con- 
tain X ;  and  one,  1  x  1,  is  independent  of  x.  Hence  the 
product  is 

(a;  +  1)  («2  +  a;  +  1)  =  ar^  +  2  ar^  -f  2  a;  + 1. 


60  MULTIPLICATION. 

EXAMPLES    XIV. 
Multiply 

1.  x  +  2yhj  x-2y.  11.    2  ?/ +  56  by  3?/ -  4  6. 

2.  a-Sbhj  a  +  Sb.  12.   3  m2  -  1  by  Sm^  +  1. 

Z.   2x  +  Syhy  Sx-2y.  U.  2m^  +  6n^hy2  m^  -  5  ii^. 

4.  5  a  +  4  6  by  a  -  6.  14.  a  +  ^  6  by  a  -  ^  6. 

5.  x+7byx  +  6.  15.  2a  +  ift  by  3a  +  i6. 

6.  X  -  7  by  X  -  6.  16.  i  a  -  1 6  by  i  a  -  i  6. 

7.  ir  +  7  by  a;  -  6.  17.  x2  +  x  +  1  by  cc  -  1. 

8.  a  +  9  by  a  -  5.  18.  x2  -  x  +  1  by  cc  +  1. 

9.  2  X  -  4  by  2  X  +  6.  19.  a^  +  a6  +  6^  by  a  -  6. 
IQ.   3x-7by2x-l.  20.  a^  -  a6  +  ft^  by  a  +  6. 

21.  4  a'^  +  6  a6  +  9  62  by  2  a  -  3  6. 

22.  16^2  +  20m  +  25  g2  by  4^)  -  5  g. 

23.  x3  -  3  ax2  +  2  a^x  by  x.-\-  3  a. 

24.  a3  -  4  a26  +  6  a62  by  a2  +  4  a6. 

25.  x3  -  3  x2  +  2  X  +  1  by  x2  +  3  x  4-  2. 

26.  x^  +  x2  -  2  X  +  1  by  x2  -  X  +  2. 

27.  x2  +  x?/  +  ?/2  by  x2  -  x?/  +  ?/2. 

28.  a*  +  a262  +  6*  by  a*  -  a26^  +  6*. 

29.  2  x^  -  3  x2?/  +  2  x?/2  +  2/3  by  x2  +  3 x?/  +  2y^, 

30.  x3  -  4  x2?/  +  6  x?/2  -  3  ?/3  by  3 x2  -  4  x?/  +  53/2. 

64.  If  in  any  expression  consisting  of  several  terms 
which  contain  different  powers  of  the  same  letter,  the 
term  which  contains  the  highest  power  of  that  letter  be 
put  first  on  the  left,  the  term  which  contains  the  next 
highest  power  be  put  next,  and  so  on ;  the  terms,  if  any, 


MULTIPLICATION.  61 

which  do  not  contain  the  letter  being  put  last ;  then  the 
whole  expression  is  said  to  be  arranged  according  to 
descending  powers  of  that  letter.  Thus,  the  expression 
a^  4-  a^h  +  aW  +  W,  is  arranged  according  to  descending 
powers  of  the  letter  a.  In  like  manner  we  say  that  the 
above  expression  is  arranged  according  to  ascending  powers 
of  the  letter  b. 

65.  Although  it  is  not  necessary  to  arrange  the  terms 
either  of  the  multiplicand  or  of  the  multiplier  in  any- 
particular  order,  it  will  be  found  convenient  to  arrange 
both  expressions  either  according  to  descending  or  accord- 
ing to  ascending  powers  of  the  same  letter  :  some  trouble- 
in  the  arrangement  of  the  different  sets  of  like  terms  of 
the  product  in  vertical  columns  will  thus  be  avoided. 
Hence,  before  beginning  to  find  the  product  of  two  ex- 
pressions, it  is  often  desirable  to  rearrange  the  terms. 

66.  A  term  which  is  the  product  of  n  letters  is  said 
to  be  of  n  dimensions,  or  of  the  nth  degree.  Numerical 
factors  are  not  to  be  counted  in  reckoning  the  number  of 
dimensions.  Thus  abc  is  of  three  dimensions,  or  of  the 
third  degree ;  and  5  a^b\  that  is  5  aabbc,  is  of  five  dimen- 
sions, or  of  the  fifth  degree. 

67.  The  degree  of  an  expression  is  the  degree  of  that  term 
which  is  of  highest  dimensions. 

•  In  estimating  the  degree  of  a  term,  or  of  an  expression, 
we  sometimes  take  into  account  only  a  particular  letter, 
or  particular  letters  :  thus  we  say  that  ax^  +  6a;  +  c  is  of 
the  second  degree  in  x,  or  is  a  quadratic  expression  in  x ;  also 
that  aoc^y  +  bxy^  is  of  the  second  degree  in  x,  and  of  the 
third  degree  in  x  and  y. 


62  MULTIPLICATION. 

When  all  the  terms  of  an  expression  are  of  the  same 
dimensions,  the  expression  is  said  to  be  homogeneous. 
Thus  aj^  +  3a-6  —  56^  is  a  homogeneous  expression,  every 
term  being  of  three  dimensions  ;  and  ax^  +  hcxy  -\-  dy^  is 
a  homogeneous  expression  in  x  and  y. 

EXAMPLES  XV. 

Arrange  the  following  expressions  according  to  descending 
powers  of  a : 

1.  a^  +  6^-2  ab.  3.    a^  +  S^  +  a%  +  ab^. 

2.  2-4a2  +  5a-6a8.  4.   5^2  _  4  _  6a3  _  2a. 

5.  aS  +  &3  +  c8  -  3  abc. 

6.  a^  +  53  _|_  c3  -(-  a^b  +  ab^  +  a^c  +  ac2  +  62c  +  bc\ 

7.  What  are  the  degrees  of  the  above  expressions,  and  which 
are  homogeneous  expressions  ? 

Bracket  together  the  different  powers  of  x  in  each  of  the  fol- 
lowing expressions : 

8.  x"^  -\-  ax  +  bx  +  ab.  9.   x"^  —  ax  —  bx  —  ab. 

10.  x^  +  ax"^  +  bx^  +  ca;2  -i-  hex  +  cax  +  abx  +  abc. 

11.  asc^  +  6aj2  +  ca;2  -f  bcx  +  caaj  +  a&a;  +  abc. 

12.  a:.2(|/-;2)+?/2(;2-a:)+;3:2(^_y). 

13.  Simplify  (Jb  +  c  -  a)x  +  (c  +  a  -  &)a:  -\-{a  +  h  —  c)x. 

14.  Simplify  (&  —  c)x+  (c  —  «)cc  +  (a  —  6)a;. 

15.  Simplify  {(6  -  a)x  +  (c  -  d)?/}  +  {(a  +  6)a;  +  (c  +  <?)«/}.     ' 

68.  Product  of  Homogeneous  Expressions.  The  product  of 
any  two  homogeneous  expressions  must  be  homogeneous ; 
for  the  different  terms  of  the  product  are  obtained  by 
multiplying  any  term  of  the  multiplicand  by  any  term 


MTJLTIPLICATION.  63 

of  the  multiplier,  and  the  number  of  dimensions  in  the 
product  of  any  two  monomial  quantities  is  clearly  the 
sum  of  the  number  of  dimensions  in  the  separate  quan- 
tities ;  hence  if  all  the  terms  of  the  multiplicand  are  of 
the  same  degree,  as  also  all  the  terms  of  the  multi- 
plier, it  follows  that  all  the  terms  of  the  product  are  of 
the  same  degree. 

When  two  expressions  which  are  to  be  multiplied  are 
homogeneous,  that  fact  should  in  all  cases  be  noticed  by 
a  student ;  and  if  the  product  obtained  is  not  homo- 
geneous, it  is  at  once  seen  that  there  is  an  error. 

69.  We  now  return  to  the  three  important  cases  of 
multiplication  considered  in  Art.  60,  namely, 

{a-\-by  =  a\+2ah-\-b^      .     .     .        (i.) 

(a-by  =  a'-2ab-\-b^     .     .     .      (ii.) 
(a  +  6)(a-6)  =  a2-62  ^     ^     (^^^ ) 

Def.  A  general  result  expressed  by  means  of  symbols 
is  called  a  formula. 

Since  the  laws  from  which  the  above  formulae  were 
deduced  were  proved  to  be  true  for  all  algebraical  quan- 
tities whatever,  we  may  substitute  for  a  and  for  b  any 
other  algebraical  quantities,  or  algebraical  expressions, 
and  the  results  will  still  hold  good. 

We  give  some  examples  of  results  obtained  by  substi- 
tution in  the  above  formulae. 

In  the  first  place,  let  us  put  —b  in  the  place,  of  b 
in  (i.) ;  we  then  have 

5a-f-(-6)J2  =  a2^2a(-6)-f(-6)2-j 


64  MULTIPLICATION. 

that  is,  (a-by  =  a^-  ^ab  +  b\ 

Thus  (ii.)  is  readily  seen  to  be  included  in  (i.) 

Now  put  6  +  c  in  the  place  of  b  in  (i.)  ;  we  then  have 

f  a  +  (&  +  c)  p  =  a^  +  2  a(6  +  c)  +  (6  +  c) 2 ; 
.-.  \a  +  b  +  cl^  =  a'-j-2ab  +  2ac  +  b^  +  2bc-^(^. 
Thus  {a+b-{-c\'=a^-\-b'+c'-^2ab-^2ac+2bc   .    (iv.) 
Now  put  —  c  for  c  in  (iv.),  and  we  have 

ja  +  6  +  (-c)P=a2  +  62  +  (_c)2  +  2a5  +  2a(-c) 
+  26(-c); 
...  (^a  +  b-cy  =  a^  +  b'  +  c'  +  2ab-2ac-2 be. 

Again  put  6  +  c  in  the  place  of  b  in  (iii.) ;  we  then 
have 

la-\-{b-{-c)]la-{b-\-c)\==a'-{b  +  cy. 

The  following  are  additional  examples  of  products 
which  can  be  written  down  at  once : 

(a2  +  52)  («2  _  ^2)  ^  (^2)2  _  (52)2  =  ^4  _  ^4^ 

(a-  6  +  c)(a  +  5  -c)  =  {a-{b-c)}{a-h(ib  -  c)} 
=  ^2  -  (6  -  c)2  =  a2  _  (52  _  2  5c  +  c^)  =  ^2  _  52  _|.  2  5c  -  c^ 
(a2  +  a5  +  52)  (a2  -  ab  +  52)  =  (a2  +  52  +  ab)  (^2  +  52  -  a5) 
=  (a2  +  52)2  _  ^252  .^  a4  +  2  ^252  +  54  _  ^252  =  ^4  4.  ^252  +  54.. 

70.  Square  of  Any  Multinomial  Expression.  We  have  found 
in  the  preceding  article,  and  also  in  Art.  62,  the  square 
of  the  sum  of  three  algebraical  quantities ;  and  we  can 
by  the  same  methods  obtain  the  square  of  the  sum  of 
more  than  three  quantities.  The  square  of  the  sum 
of  any  number  of  algebraical  quantities  may  also  be 
found  in  the  following  manner. 


MULTIPLICATION.  65 

Suppose  we  wish  to  find 

We  know  that  the  product  of  any  two  algebraical 
expressions  is  equal  to  the  sum  of  the  partial  products 
obtained  by  multiplying  every  term  of  the  multiplicand 
by  every  term  of  the  multiplier.  If  we  multiply  the 
term  a  of  the  multiplicand  by  the  term  a  of  the  multi- 
plier, we  obtain  the  term  a^  of  the  product :  we  similarly 
obtain  the  terms  b^,  c^,  etc.  We  can  multiply  any  term, 
say  b,  of  the  multiplicand  by  any  different  term,  say  d, 
of  the* multiplier;  and  we  thus  obtain  the  term  bd  of  the 
product.  But  we  also  obtain  the  term  bd  of  the  product 
by  multiplying  the  term  d  of  the  multiplicand  by  the 
term  b  of  the  multiplier,  and  we  can  obtain  the  term 
bd  in  no  other  way,  so  that  every  such  term  as  bd,  in 
which  the  letters  are  different,  occurs  twice  in  the 
product.  The  required  product  is  therefore  the  sum  of 
the  squares  of  all  the  quantities  a,  b,  c,  d,  etc.,  together 
with  twice  the  product  of  every  pair. 

Thus,  the  square  of  the  sum  of  any  mimber  of  algebraical 
quantities  is  eq^ial  to  the  sum  of  their  squares  together  with 
twice  the  product  of  every  pair. 

Ex.  1.  Find  the  square  oi  a  +  b  -\-  c.  The  squares  of  the  sepa- 
rate terms  are  a^,  6"^,  c'K  The  products  of  the  different  pairs  of 
terms"  are  ab,  ac,  and  be. 

Hence  (a  +  6  +  c)2  =  a^  +  62  +  c2  +  2  a6  +  2  ac  +  2  be. 

Ex.  3.   Find  the  square  of  a  +  2b  —  Sc. 
The  required  square 
=  a2  +  (2  6)2^(-3c)2  +  2a(2  6)-|-2a(-3c)-i-2(2  6)(-3c) 

=  a2-f  4&2-f  9c2+4a6-6ac-  12  6c. 


66  MULTIPLICATION. 

Ex.  3.   Find  («  _  &  +  c  -  dy. 

(a  -  &  +  c  -  (?)2  =  a2  ^_(_  5)2  +  c2  +  (-  <^)2  +  2  a(-  &)  +  2  ac 

+  2a(-d)+2(-&)c  +  2(-6)(-d)+2c(-d) 
=  a2  4-  62  4.  c2  +  d2  -  2  a6  +  2  «c  -  2  ad  -  2  6c  ^-  2  &(Z  -  2  ccZ. 

After  some  practice  the  intermediate  steps  can  be  omitted  and 
the  final  result  written  down  at  once.  To  ensure  taking  twice  the 
product  of  every  pair,  it  is  best  to  take  twice  the  product  of  each 
term  and  of  every  term  YflaXah.  follows  it. 

EXAMPLES  XVI. 
Write  down  the  squares  of  the  following  expressions : 

1.  2  a  +  &.  b.    a^-bah.  9.    a  -  b  -  c. 

2.  4a +  36.  6.   2a2-3a6.  10.   2a  +  25-c. 

3.  Sa-b.  7.    -Sxy  +  2y^.  11.  4a  +  26-3c. 

4.  5a -66.  8.  4^2-7^2.  12.   2a -56 -3c. 

13.  ic2+ic  +  l.  19.  3a +  26- 4c +  d:. 

14.  a;2-a:  +  l.  20.  5a  +  6  -  4c  -  3c?. 

15.  x'^-xy  +  2/2.  21.  x^  +  x^  +  X  +  1. 
■  16.  x^  +  x^y^  +  2/4.  22.  x^-x^-\-x-l. 

17.  a-b-c  +  d.  23.  2x^  -  x^y  +  xy^  -  Sy^ 

18.  2a-26-3c  +  3d!.  24.   2x^  -  x'^y +  2xy^  -  yK 
Multiply 

25.  X  -  y  +  z  hj  X  -  y  -  z. 

26.  x-2y-h4:zhjx-2y  —  4z. 

27.  Sx-y  -6zhj  Sx  +  y  -bz. 

28.  -x  +  2y-Sz\)yx-\-2y-3z. 

29.  x'^  -\-  xy  +  y2  by  x'^  —  xy  -{■  y^. 

30.  3x2  -  icy  +  2y2  by  3x2  +  0;?/ +  2?/2. 


MULTIPLICATION. 


67 


31.  x^-x-\-7\)y  x^-x-l. 

32.  2 x2  -  3 X  +  7  by  2  x2  +  3x  +  7. 

33.  a  +  6  +  c  +  rf  by  a  +  6-c-d 

34.  2a-36  +  2c-4dby2a-36-2c  +  4(2. 

35.  2a  +  36  +  c-2fZby2a-3&-c-2d. 

36.  a  -  3  6  -  4  c  +  <i  by  a  +  3  6  -  4  c  -  d. 

71.   Continued    Products.      The    continued    product  of 
several   algebraical   expressions  is  obtained   by  finding 
;he  product  of  any  two  of  the  expressions,  and  then 
.ultiplyiug  this  product  by  the  third  expression. 

Ex.  1.   Find  the  continued  product  (x  +  a)  (x  4-  b)  (x  +  c). 
The  process  is  written  as  under ; 


x  +  a 

x  +  b 

x^+ax 

bx-\-ab 

sc^  +  (a  +  b)x  +  ab 

x  +  c 

a^  +  (a  +  6)x2  +  a&x 

cx2  +  c(a  +  b)x 

+  abc 

x8  +  (o  +  b'+  c)x2  +  (a6  -\-  ac+  bc)x  +  abc 

If  a -b  =  c,  we  have  (x  +  a)^  =  x^  +  3 ax^  +  'Sa^x  +  «». 

The  above  result  is  arranged  in  a  way  which  is  frequently  re- 
quired, namely  according  to  powers  of  x,  and  all  the  terms  which 
contain  the  same  powers  of  x  are  collected  together. 

Ex.  2.  Find  the  continued  product  of  x^  +  a^,  x  +  a,  and  x  —  a. 
The  factors  can  be  taken  in  any  order ;  hence  the  required  prod- 
uct 

=  (x  -  a)  (x  -f  a)  (X-  -f  a"-)  =  (x2  -  a^)  (x'-^  +  a^)  =  x*  -  a*. 


68  MULTIPLICATION. 

Ex.  3.   Find  (x  -  ay(x  +  ay. 

Since  factors  can  be  taken  in  any  order, 

(x  -  a)%x  +  a)2  =(x  -  a)(x  +a)(x  -a)(x  +  a) 
={(x-a)(x  +  a)}2 
=  (a;2  -  a^y  =  x^-2  aH'^  +  a*. 

72.  Powers  of  a  Binomial.  We  shall  in  a  succeeding 
chapter  show  how  to  write  down  any  power  of  a  binomial 
expression,  by  a  formula  which  is  called  the  Binomial 
Theorem.  The  square  and  the  cube  of  a  binomial  should 
however  be  learnt  at  this  stage.     These  are  given  by 

and  (a  ±  6)^  =  a«  ±  ^  a% -\- 3  aV  ±  b\ 

EXAMPLES  XVII. 
Multiply 

1.  3  a2  +  a6  -  62  by  a2  _  2  a6  -  3  b'\ 

2.  x^-xy  -3 ?/2  hy  Sx^  +  xy  -  y^. 

3.  6x^  -4:X^y  +  S xy^  hy  x'^  +  4:xy+6 y^. 

4.  x^  -7 x'^y  +  Sxy"^  by  a^*  +  7 x^y  -  3  x'^y^. 

5.  3a;3_7x2  +  5ic-3by  2cc3  + 7ic2_5a;  +  4. 

6.  a^  +  Sa^-7a+Q\iySa^-a'^-h6a-4. 

7.  -|ic2  -  ^xy  +  ?/2  by  iic2  -  1  a;?/  -  y2. 

8.  ^x2-fa:?/  +  12?/2by  12a;2  +  |i«y-|y2. 

9.  a2  +  62  +  c2  -  6c  -  ca  -  a&  by  a  +  &  +  c. 

10.  4  a2  +  9  62  +  c2  -  3  6c  -  2  ca  -  6  a6  by  2  a  +  3  6  +  c. 

11.  a^  -{-b^  +  c^  +  be -\-  ca  -  ab  hy  a  +  b  -  c. 

12.  9  a2  +  62  +  9  c2  +  3  6c  -  9  ca  +  3  a6  by  3  a  -  6  +  3  c. 


MULTIPLICATION.  69 

13.  x'^  +  xy  +  fp-  by  -f  -xy  ^  x^. 

14.  a2  -  2  a6  +  4  6-2  by  4  62  +  2  a6  +  a\ 

15.  ax  4-  a2x2  +  aH^  by  a^^-ax  +  1. 

16.  (k2_|.  1^  ic+  1,  anda;-!. 

17.  a2  ^_  /c2^  05  _^  X,  and  «  —  x. 

18.  a;2_|.4y2^  a;  +  2y,  anda;-2y. 

19.  9  «2  4-  25  2/2,  3tc  +  6  y,  and  3  X  -  5  y. 

20.  x*  4-  y*,  a;'^  +  y^,  x  +  y,  and  x  -  y. 

21.  (a2  +  62)2^  («  4.  6)2^  and  (a  -  6)2. 

22.  (x2  +  X  +  1)2  and  (x2  -  x  +  1)2. 

23.  x2  -  xy  +  2/2,  x^  -{-xy  +  y^,  and  x*  -x2y2  4-  y*. 

24.  a*  -  a262  4-  54^  ^2  4-  a&  +  b%  and  a2  -  a6  +  62. 

25.  x2  —  ax  +  a2,  x2  4-  ax  +  a2,  a  +  x,  and  a  —  x. 

26.  X  +  2/  4-  2;,  -  X  +  y  +  5r,  X  —  y  +  5r,  and  x  +  y  -  sr. 
Find  the  following  cubes  : 

27.  (2  a  +  3  6)8.  29.   (3  a  -  2  6)8.  31.    (a  -  6  +c)8. 

28.  (2a -3  6)8.  30.   (^a  +  b  +  cy.         82.    (a  -  6  -  c)*. 

33.  Show  that 

(i.)  (2x  +  l)2  +  (x- 1)2  =  4x24.(0:4. 1)24.1. 
(ii.)  (2x  +  1)2  4- (X  +  2)2  =(x  -  2)2  4-  4x(x  +  3)4-1. 
(iii.)  (x2  4-  X  +  1)2  +  (x2  -  X  +  1)2  =  2  (x4  +  3 x2  4-  1). 

34.  Show  that 

a8  -  68  =(a  -  6)(a2  4-  ab  +62)  =  (a  -  6)3  4-  3a6  (a  -  6) 

=  (a  +  6)3  -  3  a6(a  +  6)  -  2  68. 

35.  Show  that 

(X  +  y)(x  +  z)-\-(y  +  z)(y  +  x)  +  (2r  +  x)(^  +  y)-(x  +  y  4-  z^ 

=  yz  +  zx-\-  xy. 


70 


MULTIPLICATION. 


36.  Show  that 
(6  -  cy+ic  -  ay  +  (a  -  by  -  3(6  -  c)(c  -  a)(a  -b)  =  0. 

37.  Simplify 

ix-]-y  +  zy-(-x  +  y  +  zy-\-(ix-y  +  zy-(x  +  y-  zy. 

38.  Simplify 

{x^y^z)  (a;2  +  1/2  +  ^2)  _  yz{y  +  2;)  -  zx{z  -^x)-  xy(x  +  y). 

39.  Show  that 

(6  -f  c) (c  +  a) (a  +  6)  +  a6c  =  (a  +  &  +  c)  (he  +  ca  +  a&). 

40.  Show  that,  iix=2y+bz\  then  will  x^ = 8  y^ + 125 z^ 4-  30  xi/0. 

41.  Show  that,  iix=h  +  c-2a,  y=c-\-a-2b,  and  z  =  a+b-2c; 

then  x2  4-  2/2  +  ^2  _(_  2  ?/0  +  2  ^x  +  2  a;?/  =  0. 

42.  Show  that  aH6^+c3+3(6+c)(c+«)(a+6)  =  (a+&  +  c)3. 

43.  Show  that  a2(5  +  c)2+62(c+a)2+c2(a+&)2+2a&c(a  +  &  +  c) 

=  2(6c  +  ca  +  aby. 

44.  Show  that  (a  +  6)3  +  3  c(a  +  6)2  +  3  c2(a  +  6)  +  c^ 

=  (6  +  cy  +  3a(6  +  c)2  +  3a2(6  +  c)-ha^. 

45.  Show  that 

{(a  +  6)^2  -  (a2  +  62)x  +  a^  +  ^sj  ^(^a  -  6)0^2  _  (^2  _^i^x  +  a^-  b^ 
=  (a2  -  62)x*  -  2(a3  -  63)x3  +  3(a*  -  b^)x^  -  2Qa^  -  b^)x  +  a^-  b^ 


8BB 


DIVISION.  71 


CHAPTER  V. 
Division. 

73.  In  multiplication  we  have  two  factors  given,  and 
we  have  to  find  their  product.  In  division  we  have  the 
product  and  one  factor  given,  and  we  have  to  determine 
the  other  factor,  so  that  to  divide  a  by  6  is  to  find  a 
quantity  c  such  that  b  x  c  =  a. 

74.  Since  division  is  the  inverse  operation  to  that  of 
multiplication  [Art.  9],  and  successive  multiplications 
can  be  performed  in  any  order  [Art.  52],  it  follows  that 
successive  divisions  can  be  performed  in  any  order. 

Thus  a-^b^c  =  a-^c-^b. 

It  also  follows,  from  Art.  52,  that  to  'divide  by  any 
quantities  in  succession  gives  the  same  result  as  to  divide 
at  once  by  their  product. 

Thus  a^b  ^c  =  a-i-  (&c). 

75.  Not  only  may  a  succession  of  divisions  be  per- 
formed in  any  order,  but  a  succession  of  divisions  and 
multiplications  may  be  performed  in  any  order. 

For  example,      axb-i-G  =  a-T-cxb. 
For  a  =  a  -f-  c  X  c ; 

.*.  a  xb  =  a-7-c  X  c  xb 

=  a-^cx&xc;         [by  Art.  52.] 


72  DIVISION. 

therefore,  dividing  each,  by  c,  we  have 

ax6-^c  =  a-J-cx5. 

Hence  we  get  the  same  result  if  we  divide  the  product  of 
two  quantities  by  a  third,  as  if  we  divide  one  of  the  quanti- 
ties by  the  third  and  then  multiply  by  the  other. 

76.  The  operation  of  division  is  often  indicated  by- 
placing  the  dividend  over  the  divisor  with  a  line  between 
them,  or  by  separating  the  dividend  from  the  divisor  by 

an  oblique  line  called  the  solidus :  thus  -  or  a/b  means 

When  a-i-b  is  written  in  the  fractional  form  -,  a  is 

6 


called  the  numerator,  and  b  the  denominator. 
By  putting  a  =  l  in  ax6-^c  =  a-i-cx&, 

we  have  lx6-^c  =  l-^cx6, 

that  is  b  -r-c  =  -  xb  =  b  X-, 

c  c 

so  that  to  divide  by  any  quantity  c  gives  the  same  result  as 

to  multiply  by  — 
c 

77.  Since  a^  x  a^  =  a^,  and  a^  xa^=  a^^,  we  have  con- 
versely a^ -i-a^—a^  =  a^~%  and  a^^ -r-a^=  a^°"^  =  a',  and 
similarly  in  other  cases. 

Hence,  if  one  power  of  a  letter  be  divided  by  another 
power  of  the  same  letter,  the  index  of  the  quotient  is  equal 
to  the  difference  of  the  indices  of  the  dividend  and  the 
divisor. 


DIVISION.  73 

78.  We  have  proved,  in  Art.  51,  that  a  x{—b)  =  —  ah; 
.'.   (— ab)-i-(—b)  =  a,  Sind  {—ab)-i- a  =  —  b. 

We  have  also  proved  that 
(—a)  X  {—b)  =  -\-ab,  and  {-^  a)  x  {-{-  b)  =  +  ab; 
.     .-.   (+a6)-^(— a)  =  — 6,  and  (f  a6)^(+a)  =  +  5. 

Hence,  if  the  signs  of  the  dividend  and  divisor  are 
alike,  the  sign  of  the  quotient  is  +  ;  and  if  the  signs  of 
the  dividend  and  divisor  are  unlike,  the  sign  of  the 
quotient  is  —  ;  we  therefore  have  the  same  Law  of  Signs 
in  division  as  in  multiplication. 

79.  Division  of  Monomial  Expressions.  The  results  arrived 
at  in  the  preceding  articles  will  enable  us  to  divide  any- 
one monomial  expression  by  another.  These  results 
are: 

(i.)  The  sign  of  the  quotient  is  +  when  the  signs  of 
the  dividend  and  divisor  are  alike,  and  the  sign  of  the 
quotient  is  —  when  the  signs  of  the  dividend  and  divisor 
are  different.     (Law  of  signs.) 

(ii.)  The  operations  of  multiplication  and  division 
may  be  performed  in  any  order.  (Laws  of  commutation 
and  association.) 

(iii.)  When  one  power  of  any  letter  is  divided  by 
any  smaller  power  of  the  same  letter,  the  index  of  the 
quotient  is  equal  to  the  difference  of  the  indices  of  the 
dividend  and  the  divisor.     (The  index  law  for  division.) 

Ex.  1.   Divide  18  a*b^  by  6  a^&s. 

18  a*66  ^  6  a263  =  is  ^6  x  a^  ^  a^  x  b^ -^  b^,      from  (ii.) 
=  3  a*-265-3,  from  (iii.) 

=  3a2&2. 


74  DIVISION. 

Ex.  2.    Divide  15  a%^  by  -  5  ahK 

15a366^(-5a65)  =  -l5H-5xa3-ax66H-6^   from  (i.)  and  (ii.) 
=  -3a3-i56-5^  from  (iii.) 

Ex.  3.   Divide  -  5  a'h^d^  by  -  7  a%'^(^. 

The  whole  work  of  dividing  one  monomial  expression  by  another 
can  be  performed  mentally  and  the  result  written  down  at  once. 
First  put  down  the  sign  of  the  result ;  then  divide  the  numerical 
coefficient  of  the  dividend  by  that  of  the  divisor ;  then  take  each 
letter  which  occurs  in  the  dividend  to  a  power  whose  index  is  the 
difference  of  the  indices  of  that  letter  in  the  dividend  and  the 
divisor. 

Thus  (-  5 a^&V) - (-7  a^&V)  =  +  f  a^h^,  since  c*  --  c*  =  1. 

If,  in  the  above  example,  we  had  used  the  rule  for  finding  the 
index  of  the  quotient  of  c*  ^  c*,  we  should  have  been  led  to  the  at 
present  meaningless  result  &  :  but  c*  -r-  c*  is  obviously  1. 

80.  Division  of  a  Multinomial  Expression  by  a  Monomial. 
When  a  multinomial  expression  is  divided  by  a  monomial, 
the  quotient  is  equal  to  the  sum  of  the  quotients  obtained 
by  dividing  its  separate  terms  by  that  monomial. 

Thus         {a-\-b  t\-  '")  -i-x=:  a-i-  x-\-b  -i-x-{-  "' 
To  prove  this,  multiply  by  x ;  then 

(a  +  b-\ )^x  xx  =  a-\-b-] — ; 

also  {a-i-x-{-b^x-\ — )  xx  =  a-irXXx  +  b-v-xxx-\ — 

=  «  +  &  +  •••          [Art.  57.] 
Hence        (a-\-b-\ )-i-x  =  a-^x-{-b-i-x-\ 

Ex.  1.   Divide  a^  +  2a'^  by  a. 

The  result  is  a^  ^  a  -h  2 a"^  -^  a  =  a'^  +  2a. 


DIVISION.  15 

Ex.  2.  Divide  a^x^  —  3  ax  by  ax. 

The  result  is        a^x^  -=-  ax  +  (  —  3  ax)  -4-  ax  =  ax  —  3. 

Ex.  3.   Divide  12  x^  -  5  ax2  -  2  a^x  by  3  x. 

The  result  is 
12x3--3x  +  (-5ax2)  ^  3x +(- 2a2x) --3x  =  4x2  -  f  ax  -  fa^. 

EXAMPLES  XVIII. 
Divide 

1.  10  a  by  -  5  a.  10.  25  a^b^c^  by  -  5  a^bC*. 

2.  -  10  6  by  2  6  .  11.    -  27  a^ftHc*  by  -  6  a^bK 
8.    -2xby-3x.  12.  o6cd  by -a&. 

4.  a2  by  -  a.  13.  a^¥c*(f  by  -  2  a62c#. 

6.  8  a6  by  -  2  b.  14.  8  a^x^y'^z  by  6  ax^y*. 

6.  -  4  a6  by  3  a.  15.  -  3  a^b^cx'^y^  by  -  2  a&cx=V^- 

7.  12  a268  by  -  a&2.  16.  3  x^  -  5  ax  by  x. 

8.  -6 x2y8  by  -  4 xy.  17.  by*  -Qy^hj  -  y^. 

9.  -3xVl3y2xy2.  18.  4a6  -  Sa^  +  2a*  by  a*. 
X 19.  12  a8  +  9  a*  -  6  a5  by  -  3  a2. 

N  20.    15  a*66  _  7  a^b''  +  9  a264  by  -  3  a'^b\ 
V  21.   5(x  -  1)2  -  3  a(x  -  1)  by  (x  -  1). 

22.    12  a6(x  -ay -6  a*(x  -  a)^  +  3  a^(x  -  ay  by  a^(x  -  ay. 

81.  Division  by  a  Multinomial  Expression.  We  have  nov^ 
to  consider  the  most  general  case  of  division,  namely, 
the  division  of  one  multinomial  expression  by  another. 

The  object  of  division  is  to  find  by  what  algebraical 
expression  the  divisor  must  be  multiplied  to  produce 
the  dividend. 


76  DIVISION. 

Consider,  for  example,  the  division  of 

Sa^-\-Sa'b-\-4.ab^-\-b^hj2a  +  b. 

The  dividend  and  the  divisor  must  first  be  both 
arranged  according  to  descending  (or  ascending)  powers 
of  some  common  letter,  and  the  terms  of  the  quotient 
will  be  found  separately  in  the  same  order.  In  the 
present  case  the  arrangement  is  according  to  descending 
powers  of  the  letter  a.  Now  the  term  of  highest  degree 
in  a  in  the  dividend,  namely,  8  a^,  must  be  the  product 
of  the  terms  of  highest  degree  in  a  in  the  divisor  and 
the  quotient ;  hence  the  Jii^st  term  of  the  quotient  must 
be  8  a^  -7-  2  a  =  4  al  Now  multiply  the  divisor  by  4  a^ 
and  subtract  from  the  dividend:  we  then  have  the  re- 
mainder 4  a^6  +  4  ab"^  +  b^.  This  remainder  must  be  the 
product  of  the  divisor  by  the  other  terms  of  the  quotient ; 
and  hence  the  first  term  of  the  remainder,  namely  4  a^b, 
must  be  the  product  of  the  first  term  of  the  divisor  and 
the  next  term  of  the  quotient.  Hence  the  second  term 
of  the  quotient  is  4  a^6  -^  2  a  =  2  ab.  Now  multiply  the 
divisor  by  2ab  and  subtract  the  product  from  the  re- 
mainder :  we  then  get  the  second  remainder  2  ab^  +  61 
The  third  term  of  the  quotient  is  similarly  2  a6^  -^  2  a  =  b\ 
Multiply  the  divisor  by  b^  and  subtract  the  product  from 
2  ab^  -\-  b^,  and  there  is  no  remainder.  Since  there  is  no 
remainder  after  the  last  subtraction,  the  dividend  must 
be  equal  to  the  sum  of  the  different  quantities  which 
have  been  subtracted  from  it;  but  we  have  subtracted 
m  succession  the  divisor  multiplied  by  Aa^j  by  2ab, 
and  by  b^;  we  have  therefore  subtracted  altogether  the 
product  of  the  divisor  and  4  a^  +  2  a6  +  b\  Hence  the 
required  quotient  is  4  a'^  +  2  a6  +  b\ 


DIVISION. 


77 


The  process  is  written  in  the  following  form  : 

2a  +  6 


8a^ 

-\-Sa'b-{-4.ab'-\-b' 
-h^a'b 

4:a'b-{-4:ab'-{-b^ 
4:a'b-{-2ab^ 

2ab''  +  W 
2ab^-{-b^ 

4a24-2a6  +  62 


From  the  above  example  it  will  be  seen  that  in  order 
to  divide  one  multinomial  expression  by  another  we 
proceed  as  follows : 

(1)  Arrange  both  dividend  and  divisor  according  to 
ascending  or  both  according  to  descending  poivers  of  some 
common  letter. 

(2)  Divide  the  first  term  of  the  dividend  by  the  first  term 
of  the  divisor :  this  will  give  the  first  term  of  the  quotient. 

(3)  Multiply  the  divisor  by  the  first  term  of  the  quotie^it 
found  above,  and  subtract  the  product  from  the  dividend. 

(4)  Now  treat  the  remainder  as  a  new  dividend,  and  go 
on  repeating  the  process. 

82.   The  following  are  additional  examples : 

Ex.  1.   Divide    3 x^  -  4 x2  +  2x  -  1  by  1  -  a;. 
The  order  of  the  terms  in  the  divisor  must  first  be  changed 
to  -  X  +  1. 

3x8-3a;g |  -Zx^  +  x-l 

-  x2  +  2  X  -  1 

-  X2+X 


x-1 
x-1 


78 


DIVISION. 


Ex.  2.  Divide  a*  -a%-\-2  a^"^  -  ab^  +  b^  by  a"^  +  b\ 


a^  -a%  +  2  a%^  -  ab^  +  &* 


a2+&2 


«'-^  -  a6  +  62 


-a36  + 

a262  - 

-  ab^  +  M 

-a% 

-a63 

+ 

a%^ 

+  6* 

+ 

a'^b^ 

+  6* 

Ex.  3.   Divide  a*  +  ^^252  +  ^4  by  ^2  _  ^^  ^  52. 


a*  +  a262 

a^  -  a%  +  a262 


+  6^ 


a2  _  a6  +  62 


a2  +  a6  +  62 


+  a3?, 

+  6* 

+  a^6 

-  a262  _f.  ^53 

+  a262  - 

-  a63  +  64 

+  a262  _ 

-  ab^  +  6* 

In  the  last  example  the  terms  of  the  dividend  were  placed  apart 
in  order  that  '  like '  terms  might  be  placed  under  one  another 
without  altering  the  order  of  the  terms  according  to  descending 
powers  of  a.  The  subtractions  can,  however,  be  easily  performed 
without  putting  '  like '  terms  under  one  another ;  but  the  arrange- 
ment of  the  terms  according  to  descending  (or  ascending)  powers 
of  the  chosen  letter  should  never  be  departed  from. 


83.  Horner's  Synthetic  Division.  In  the  foregoing  process 
of  division  there  is  a  redundancy  of  work  which  may  be 
avoided  by  a  different  arrangement  of  the  successive 
steps.  Fixing  upon  the  idea  of  division  as  the  process 
that  undoes  multiplication  [Art.  9],  let  us  first  multiply 
together  the  divisor  and  quotient  of  Ex.  3  of  the  last 
article,  obtaining  the  dividend  as  their  product  and 
arranging  the  work  in  rectangular  form  as  follows: 


DIVISION. 


79^ 


D 

—  ab 


a^^ab  4-  b^ 


a^  +  a%  +  a'b' 

-a'b-a'b^-abK  .  . 
-h  oJ'b^ -^  ab^ +b'' 


a^  4-  0  +  CL-b^  +  0    +  &' 


=  Q 

=Pi 

=  P2 
=  P3 
=  P. 


For  brevity's  sake  the  several  multinomials  of  the 
diagram  will  be  referred  to  by  means  of  the  attached 
letters  D,  Q,  pi,  p^,  Psy  P- 

In  order  to  retrace  the  steps  here  indicated  and  recover 
the  quotient  Q  from  P  and  D  (the  dividend  and  (fi visor), 
regarded  as  known  or  given,  we  have  merely  to  subtract 
from  P  the  sum  of  the  two  partial  products  p^  and  p^  and 
divide  the  result  by  a? ;  for  P  was  obtained  by  adding 
together  p^,  Ps,  and  a^Q.  Or  better,  instead  of  subtracting 
P2  +  Pi  from  Pj  we  may  change  all  the  signs  in  pg  and  p^, 
add  together  —p2,  —Ps,  and  P,  and  then  divide  the 
result  by  a^,  as  before.  Now  we  can  obtain  the  succes- 
sive terms  of  —p2  and  —p^  by  multiplying  the  negatives 
of  the  last  two  terms  of  D  by  the  successive  terms  of  Q 
as  fast  as  found;  and  these  are  found,  by  addition,  as 
fast  as  the  columns  in  the  above  arrangement,  proceed- 
ing from  left  to  right,  are  completed  by  the  successive 
multiplications  here  indicated. 

In  order  to  exhibit  the  process  in  a  convenient  form, 
we  modify  the  arrangement  in  the  multiplication  process 
by  interchanging  P  and  Q,  suppressing  pi,  and  changing 
all  the  signs  in  the  lines  in  which  p2  and  p^  stand.    Thus, 


a' 
+  ab 

-^-a^b  +  a'W^aW    .  .  .  . 
.    -a'b'^ab^-b'  .  . 

.   =P 

•  =-P2 

•  =-Ps 

a^-\-ab  +b^i        0       0   .  . 

.  =Q 

80  DIVISION. 

The  successive  steps  of  the  work  are  as  follows : 

(1)  The  dividend  and  the  divisor,  with  all  its  signs 
changed  except  that  of  the  first  term,  are  written  down 
as  indicated,  arranged  according  to  descending  powers 
of  a,  the  zero  coefficients  of  o?  and  a  being  retained. 

(2)  The  first  term  a^  of  the  quotient  is  found  as  the 
quotient  of  a^  by  o?. 

(3)  The  oblique  column  ~  _  21,2  is  got  by  multi- 
plying +  ah  and  —  V^,  the  last  two  terms  of  the  divisor 
(with  signs  changed),  by  a^,  the  first  term  of  the 
quotient. 

(4)  The  second  term  ah  of  the  quotient  is  the  sum 
of  the  terms  in  the  second  column,  divided  by  o?. 

(5)  The  terms  +a^6-,  —aW  of  the  second  oblique 
column  are  the  products  of  ah,  the  second  term  of  the 
quotient,  by  +  ah  and  —  61 

(6)  The  third  terra  6^  of  the  quotient  is  the  sum  of 
the  terms  in  the  third  column,  divided  by  a^. 

(7)  The  terms  ah^,  —W  of  the  third  oblique  column 
are  the  products  of  6^,  the  third  term  of  the  quotient,  by 
+  ah  and  —  6^. 

(8)  The  terms  of  the  last  two  columns  destroy  one 
another. 

In  this  process,  since  the  quantities  sought  are  merely 
the  coefficients  of  the  different  powers  of  a  in  the 
quotient,  these  powers  of  a  may  be  omitted  until  the 
coefficients  are  found,  and  be  then  inserted  in  their 
proper  places  in  the  quotient.  The  work  then  appears 
in  the  following  abbreviated  form : 


DIVISION. 


81 


1 

-62 


1  +  0  + 6' +  0+6* 
+  6  +  62  +  6« 


1  +  6  +  62;    0      0 

Quotient :  a^  +  a6  +  6^. 

The  method  here  described  is  known  as  Horner's  syn. 
thetic  division. 

84.   Other  examples  of  the  method : 
Ex.  1.  Divide ic*-2a;2+8a5-3bya;2  4-2a;-l. 

8-3 


1 

-2 
+  1 


1  +  0-2 

-2  +  4 
+  1 


6 
2  +  3 


1-2  +  3;    0     0 

Quotient :  x^  -  2  x  +  3. 

Ex.  2.  Divide  4x^  -  G  x*  +  i  x^  -  11  x'^  -{■  1  hy  2  x^  -  Sx  -  1. 


2x2 
f3x 
+  1 


4xS-6x4  +  4x3-llx2  +  0x+l 
+  6x*  +  0      +9x2-3x 

+  2x8+0     +3x-l 


-  Quotient :     2  x^,  +  0  x^  +  3  x 


0 


0 


Observe  that  the  terms  of  the  quotient  are  here  obtained  by 
dividing  the  sums  of  the  terms  in  the  respective  columns  by  2  x^. 

85.  If  the  division  be  exact,  zeros  will  appear  at  the  right 
of  the  quotient,  in  the  diagram  of  the  synthetic  method,  and 
their  number  tvill  be  one  less  than  the  number  of  terms  in 
the  divisor. 

For,  when  the  dividend,  divisor,  and  quotient  are  made 
complete  by  inserting  all  the  zero-terms  (powers  of  the 
leading  letter  with  zero-coefficients),  the  number  of  terms 


82  DIVISION. 

in  each  is  one  greater  than  the  highest  exponent  of  its 
leading  letter,  while  the  highest  exponent  of  the  dividend 
is  the  sum  of  the  highest  exponents  of  the  divisor  and 
the  quotient.  Hence  the  number  of  terms  in  the  dividend 
is  one  less  than  the  number  of  terms  in  the  divisor  and 
quotient  combined.  But  the  last  row  of  our  diagram  has 
the  same  number  of  terms  as  the  dividend,  and  therefore 

Number  of  zeros  =  (number  of  terms  in  dividend) 
—  (number  of  terms  in  quotient) 
=  (number  of  terms  in  divisor)  —  1. 

Should  the  zeros  not  make  their  appearance,  as  re- 
quired in  exact  division,  the  algebraic  expression  which 
takes  their  place  is  called  a  remainder.  [See  Art.  89.] 
It  is  well  to  separate  this  remainder  from  the  quotient 
by  a  distinctive  mark,  a  line,  or  a  semicolon. 

86.  Note.  The  foregoing  exposition  of  the  division  process  in 
its  two  principal  forms  (Arts.  81-84)  has  employed  only  particular 
examples  for  the  purpose,  and  is,  therefore,  not  in  the  form  of 
strict  algebraic  proof.  But  in  many  such  cases  a  formal  proof  is 
not  deemed  necessary.  A  general  principle  can  sometimes  be 
recognized  through  its  application  to  a  particular  example. 

87.  The  following  formulae  are  very  important,  and 
should  be  remembered ;  they  can  easily  be  verified. 

(flj^  +  2  ax  -\- a^)  ^  {x  +  a)  =  X -{- a, 

{pi?  —  2ax-\-a'^)^{x  —  a)  =  x  —  a, 
{x^  —  a^)       -i- {x  —  a)  =  X -{-  a, 
{o?  —  o?)        -i- {x  —  a)  =  x^  -\- ax -\-  o?, 
(ar^  4-  a^)        -=r- {x  -\-  a)  —  o?  —  ax  ■\-  a^, 
{x"^  —  a'')        -=-(«  —  a)  =^-\-  ax^  +  a^x  +  a^, 
{x^  —  a^)       -^  {x -{- a)  =  V?  —  ax^ -\- a^x  —  a^. 


DIVISIOIT.  88 

EXAMPLES    XIX. 
Divide 

1.  ic2-5a;  +  6l)y  x-2.  11.   x^  -  ^y^  hy  x- ^y^. 

2.  a;2  +  5a;-24by  x  +  8.  -^2.    ^aj^  -  ^^a^  by  Jx^  -  ^a^. 

3.  x2-23x+132by  x-11.  13.  3x2  +  8x  +  4  by  6x  +  12. 

4.  a;2  -  4  X  -  77  by  X  +  7.  14.   x^  -  ^V  a;  -  1  by  4  x  +  3. 

5.  3x2-4x-4byx-2.  15.   a^  -  b^  by  a  -  b. 

6.  3x2-llx+10by3x-6.  16.   8a8  +  27  &»  by  2a  +  36. 

7.  a2  ^  62  by  a  -  b.  17.    a^x^  +  6^  by  ^x  +  by. 

B:  x2  -  9 ?/2  by  X  -  3 j/.  18.   8 a6x6-125  f  hy  2 a^x^-by'^. 

9.   x2  -  16 y2  by  X  +  4 y.  19.   x^  -  8x  -  3  by  3  -  x. 

10.   9x2-64y*by3x  +  8y2.        20.  2x8  -  6x2  +  4  by  2  -  x. 

21.  2x*-9x8  +  10x2- 6x  +  6by  3-x. 

22.  X*  +  2  x2  -  X  +  2  by  1  --  X  +  x2. 

88.   We  now  give  examples  which  require  greater  care 

in  the  arrangement  of  the  terms. 

Ex.  1.   Divide  2a2-2  62_3c2_55c-5ca-3a6  by  a-26-3c. 

Where,  as  in  tliis  example,  more  than  two  letters  are  involved, 
we  must  not  merely  arrange  the  terms  according  to  powers  of  a, 
but  h  also  is  given  the  precedence  over  c.  The  terms  are  there^ 
fore  arranged  as  under. 


2a2_ 
2a2- 

-3a6- 
-4a6- 

6ac  - 
Qac 

-262- 

-6^c- 

-3c2  a- 
2a 

-26-3c 

-f-6-l-c 

a6  + 
ab 

ac  - 

-2  62- 
-262- 

-66c- 
-36c 

-3c2 

+ 

ac 
ac 

- 

-2  6c- 
-2  6c- 

-3c2 
-3c2 

84 


DIVISION. 


Ex.  2.  Divide  a^  +  b^  +  c^  -  Sabchy  a  +  b  +  c. 

First  arrange  the  dividend  according  to  powers  of  a,  and  give 
6  precedence  over  c  througliout. 


a*  -  3  abc  +  6^  +  c^ 


«  +  6  +  c 


a^  -  ab  -  ac  +  62  -  6c  +  c^ 


-  a26- 

-  a25_ 

-a2c 
-a62 

-  3  a6c  +  63  +  c3 

—  a6c 

-  a2c  +  «62  - 
-a2c 

2  a6c  +  63  +  c8 
a6c  —  ac2 

+  a62- 
+  a62 

«6c  +  «c2 

+  63  +  c3 
+  68  +  62c 

— 

a6c  +  ac^ 
abc 

-  62c  +  C8 

-  62c  -  6c2 

ac2  +  6c2  +  c3 
ac2  +  bc"^  +  c3 

The  work  of  this  example  is  much  shorter  by  the  synthetic 
method.  Arrange  the  terms  according  to  descending  powers  of  a, 
and  retain  only  their  coefficients  and  the  term  not  containing  a. 
Thus, 


-(b  +  c) 


1        0        -  3  6c 

-C6  +  c)  +  (6  +  c)2 


+  (63  +  c3) 

-(6  +  c)(62  +  c2-6c) 


l_(6  +  c)  +  (62+ c2-6c);  0 

Hence  the  quotient  is 
a^-(b  +  c)a  +  62  +  c2  -  6c  =  a2  +  62  +  c2  -  6c  -  ca  -  ab. 

Ex.  3.  Divide  a'^(b  -  c)  +  62(c  -  a)+  c\a  -b)hY  a-b. 

Arrange  the  terms  according  to  descending  powers  of  a,  and 
retain  only  their  coefficients  and  the  term  not  containing  a.    Thus, 


1 

+  6 


(5  _  c)  -  (62  -  c2)  +  (62c  -  6c2) 
4- (6  -c)6+(6c2-62c) 


(6-c)  +  (c2-6c);  0 

Hence  the  quotient  is  (6  -  c)a  +  (c  -  6)c. 


DIVISION.  85 

89.  In  the  process  of  division  as  described  in  Art.  81, 
the  dividend  and  divisor  being  arranged  according  to  de- 
scending powers  of  a,  we  subtract  from  the  dividend,  at 
the  first  subtraction,  an  expression  such  that  the  term  in 
it  which  contains  the  highest  power  of  a  is  the  same  as 
the  term  in  the  dividend  which  contains  the  highest 
power  of  a ;  hence  it  follows  that  the  highest  power  of  a 
contained  in  the  remainder  after  the  first  subtraction  will 
be  less  than  the  highest  power  of  a  contained  in  the  divi- 
dend. Similarly,  the  highest  power  of  a  contained  in 
every  remainder  will  be  less  than  in  the  preceding  re- 
mainder; and  hence  by  proceeding  far  enough,  we  must 
come  to  a  stage  where  there  is  no  remainder,  or  else 
where  there  is  a  remainder  the  highest  power  of  a  in 
which  is  less  than  the  highest  power  of  a  in  the  divisor, 
and  in  this  latter  case  the  division  cannot  be  exactly 
performed. 

It  is  convenient  to  extend  the  definition  of  division  to 
the  following :  To  divide  Aby  B  is  to  find  an  algebraical 
expression  C  such  that  B  x  C  ^s  either  equal  to  A,  or  differs 
from  A  by  an  expression  which  is  of  lower  degree,  in  some 
particular  letter,  than  the  divisor  B. 

For  example,  if  we  divide  x'^  +  S  ax  -\-  a^  hy  x  +  a,  we  have 

X  \x^  +  Sax  +  a^ 
—  a  \       —ax  —  2a'^ 


X  +2a;  -a\ 
Thus 

(x2  +  3  aa;  +  a2)  -J.  (x  +  a)  =  «  +  2  a,  with  remainder  -  «'.  [Art.  84.] 

We  may  express  the  above  in  tJie  following  form : 

a;2  +  3ax  +  a"-2=(x-|-2a)(x  +  a)-a^. 


86 


DIVISION. 


We  have  also,   by  arranging  the  dividend    and    the    divisor 
differently, 


-  a;a  -  2  a;2 


a   +  2  X ;  —  aj2^ 


and  the  remainder  is  now 


Hence,  when  there  is  a  remainder,  a  change  in  the  order  of  the 
dividend  and  the  divisor  leads  to  a  result  of  different  form.  This 
is,  however,  what  might  be  expected  considering  that  in  the  first 
case  we  find  what  the  divisor  must  be  multiplied  by,  in  order  to 
agree  with  the  dividend  so  far  as  the  terms  which  contain  x  are 
concerned  ;  and  in  the  second  case  we  find  what  the  divisor  must 
be  multiplied  by,  in  order  to  agree  with  the  dividend  so  far  as  the 
terms  which  contain  a  are  concerned. 

When,  therefore,  we  have  to  divide  one  expression  by  another, 
both  expressions  being  arranged  in  the  same  way,  it  must  be 
understood  that  this  arrangement  is  to  be  adhered  to. 


Divide 
1. 


10. 
U. 


EXAMPLES  XX. 

a;4  +  ic2  +  1  by  ic2  +  X  +  1. 

1  +  X*  +  x8  by  1  -  x2  +  X*. 
a;4  ^_  4  a;2  4- 16  by  x2  +  2  X  +  4. 
16x4  +  36x2  +  81  by  4 x2  -  6x  +  9. 
x*  +  2  x2  +  X  +  2  by  x2  +  X  +  1. 

2  ic*  -  x3  +  2  x2  +  1  by  1  -  X  +  x2. 
x5-4x3  +  2x-4by  2-x. 
x6  _  3  x2  -  52  by  2  -  X. 
I_x  +  5x3-3x4by  l-2x  +  3x2. 
a;6  _|.  2  x5  -  4  x*  -  2  x3  +  12  x2  -  2  X  -  1  by  x2  +  2  X  -1. 
l-4x8  +  3x*by  (l-x)2. 


DIVISION.  87 

12.  l-6a^  +  4a^by  (x-iy. 

13.  35  +  4x-16a;2  +  i9a:3-6x*by  7  +  6x-Sx^. 

14.  3a^  -  12x*  +  17 x3  -  19x2  +  1  by  (1  -  x)3. 

15.  x5  -  5a;2  +  5 X  -  1  by  1  +  x2  -  2x. 

16.  16x6 -97x*- 84x2  +  77x  +  8by  4x2+  11  x  +  1. 

17.  1  +  2  x5  +  x6  +  2  x^  by  1  +  X  +  x2. 

18.  xio  +  x5  +  1  by  x2  +  X  +  1. 

19.  6  x6  -  7  x^y  +  x3?/2  +  20  x2y3  -  22  xy*  +  8  y5  by  2  x2  -  3  xy  +  4  y\ 

20.  8?/5  _  22xy^  +  20x2^/3  +  x^  -  7x4i/  +  6x5  by  4y2  _  Zxy  +  2x2. 

21.  x2  —  2/2  _  a:^;  +  2/2  by  X  -  y. 

22.  x2  -  j/2  _|.  2  X2  -  2  ?/0  by  X  -  2/. 

23.  a2  -  62  +  ac  -  6c  by  a  +  6  +  c. 

24.  x2  -  2/2  -  ;s2  _|_  2  2/2;  by  2/  -  2;  -  X. 
26.  rt2  _  4  52  _  e2  _  4  6c  by  2  6  +  c  -  a. 

26.  a2  +  4  62  +  9c2  -  12  6c  -  6 ca  +  4  a6  by  3 c  -  2  6  -  a. 

27.  2  a2  -  2  62  -  2  c2  -  4  6c  +  3  ca  +  3  a6  by  a  +  2  6  +  2  c. 

28.  3  a2  -  16  62  -  4  c2  +  20  6c  -  11  ca  +  8  a6  by  a  +  4  6  -  4c. 

29.  x3  +  ?/8  _  ;53  ^  3  ;;c?/s  by  X  +  2/  -  0. 

30.  8x8  -  2/^  +  «3  +  6x2/0  by  2/  -  2  -  2x. 

31.  27a3_863  +  c8+ 18a6cby  3a-26 +  c. 

32.  a^  +  3a26  +  3a62  +  6^  +  c^  by  a  +  6  +  c. 

33.  a-*  +  6*  +  c*  -  2  62c2  -  2 c2a2  -  2^262  by  a  +  6  +  c. 

34.  (x+l)3+2/3by  X+2/+I.  35.    {x^-yzy+8fz^hyx^-^yz. 

36.  (x2  -  2  yzy -27  y^z^  by  x2  -  6  yz. 

37.  9  a2  +  6  a6  +  62  -  4  c2  -  4  c<Z  -  d2  by  3  a  +  6  -  2  c  -  (i. 

38.  x3+(4a6-62)x-(a-2  6)(a2  +  3  62)  byx-a  +  26. 


88  MISCELLANEOUS  EXAMPLES   I. 

39.  a3(6  -  c)  +  h^{c  -  a)+  c^{a  -  h)  by 

a\h  -  c)  +  62(c  -  a)  +  c%a  -  &). 

40.  a^(h  -  c)+  6*(c  -  a)+  c\a  -  h)  by 

a2(6  -  c)  +  62(c  _  a)  +  c2(a  -  h). 

MISCELLANEOUS   EXAMPLES  I. 

A.  1.   Find  the  value  of  {y  -  zf  +  (s  -  xy  +  (x  -  yY  when 
ic  =  —  1,  ?/  =  0,  and  z  —  \. 

2.  Add    together    3  a2  _  2  ca  -  2  a6,    2  62  +  3  5c  +  3  6a,    and 
c2  -  2  ac  -  2  6c. 

3.  Show  that 

a3  4.  53  4.  c8  _  3  a6c  -  a(a'^  -  be)  -  b(b^  -  ca)  -  cif'  -  ab)  =  0. 

4.  Multiply  2x2-^a;  +  |by^x  +  3, 
and                 ioc;^  +  ixy  +  y^hy  \x^  —  ^xy  +  y^. 

6.   Simplify  {x  +  1) (x  +  2) (x  +  3)  -  (x  -  1) (x  -  2) (x  -  3). 
6.    Divide  2  -  12x5  +  lOx^  by  1  -  2 x  +  x2. 

B.  1.   Find  the  value  of  c2  +  6(c  +  a),  and  of  \/2  be  —  a,  when 
a  =  II,  6  =  3,  and  e  =  6. 

2.   Subtract  5  a*  -  3  a^ft  4.  4  ^252  from  5  6^  -  3  6%  +  4  62a2. 


3.  Simplify      x  -(I  -  1  -  x),  3 x  -  7  -  4x, 
and  6  —  2a  —  {c  —  a—  (6  —  a  +  c)}. 

4.  Multiply  9  x2  —  1  by  x2  +  I,  and  a  +  6  +  cbya  +  6  —  c. 

6.   Show  that 
2(a2  +  62  +  c2  -  6c  -  ca  -  a6)  =  (6  -  c)2  +  (c  -  a)2  +  (a  -  6)2. 

6.    Divide  x^  -  3  x2  +  3  x  +  ?/3  -  1  by  x  +  ?/  -  1. 
C.  1.   Find  the  numerical  value  of 


Va2  +  2  6c  ,  \/62  -{-ea,  Vc2  +  a6 

1 1  } 

a  6  c 

when  a  =  4,  6  =  3,  and  c  =  —  2. 


MISCELLANEOUS   EXAMPLES  I.  89 

2.  Add  together  3a'^b-6  ab^+7  b^  and  2  a^  _  i  a^ft  +  5  a62  _  4  53, 

3.  Multiply  ax2  —  X  +  a  by  ax2  +  X  +  a,  and  find  the  square  of 

2  a  +  6  -  3  c. 

4.  Simplify  (x+yy-(x-\-y)(x  -  y)- {x(2y-x)-y(2x-y)}. 

5.  The  product  of  two  algebraical  expressions  is 

x°  +  x^y  +  x*2/2  _  x^y^  +  y^, 
and  one  of  them  is  x^  +  xy  +  y^ ;  what  is  the  other  ? 

6.  Prove 

(i.)  a(6  -c)+  5(c-a)  +  c(a-  6)=0, 
(ii.)  a2(6-c)  +  &2(c-a)  +  c2(a-6)  +  (6-c)(c-a)(a-6)  =  0. 

D.  1.  If  a  =  1,  6=2,  and  c=  -3,  find  the  value  of  6a  +  36+4c, 
andof  aS  +  b^-^c^-Sabc. 

2.  Simplify  3{x  -  2(y  -  z)}  -[^y  +  {2y  -(z  -  x)}]. 

3.  What  must  be  added  to  {a  +  b  +  c)^  that  the  sum  may  be 
(a-b-cy? 

4.  Multiply  a^  +  6^  by  a  —  6,  and  divide  the  result  by  a  +  6. 

6.   Divide  x^  +  ^^  by  x  4-  2,  and  from  the  result  write  down  the 
quotient  when  (x  +  yy  +  z^  is  divided  hy  x  -^  y  +  z. 

6.  Prove 

(i-)    (X  +  y)ix  -y)  +  (y+  z)(y  -  z)  +  {z  +  x)iz  -  x)  =  0, 
(ii.)    {x-yy+{y-zY-\-{z-xy 

=  2(x  -  y)(x  -  z)+2(y  -z)(y-  x)+  2(2  -  x)(z  -  y). 

E.  1.  Find  the  value  of 

2  V(a2  +  62),^  ^(a2  +  52  -  c^  +  6(f^) 
d  —  C  +  b  —  a 
when  a  =  3,  6  =  4,  c  =  6,  and  d  =  6. 

2.  Simplify  4  x  -  (y  -  x)  -  3 {2  ?/  -  3(x  +  y)}  and 
2x-3y-4(x-2y)+5{3x-2(x-  y)}, 
and  multiply  the  two  results  together. 


90  MISCELLANEOUS   EXAMPLES  I. 

3.  Show  that  {a  +  by  =  a^  +  b^  +  S  ab(a  +  h),  and  verify  the 
result  when  a  =  1  and  b  =  —  2. 

4.  Find  the  coefficient  of  x^  in  the  product  of  x^  —  (^a—  b)x  —  ab 
and  x2  +  (a  +  b)x  +  ab. 

6.  Divide  the  difference  of  (2  a  +  3  by  and  (3  a  +  2  by  by 
a  +  6 ;  divide  also  the  sum  of  (2  a  —  3  6)3  and  (3  a  —  2  &)3  by  a  —  &. 

6.    Show  that 
(a  +  b  +  c  - d)(a  +  b  -  c  -\-  d)  +  (a -  b  +  c  +  d)(-  a  +  b -h  c -\-  d) 

=  4(a6  +  cd). 

F.  1.   Find  the  value  of 

(g  +  6)  (c  +  <?)-(&  +  c)  (d  +  g) 
g6  +  6c  —  cd  —  (?g 

when  g  =  3,  6  =  4,  c  =  5,  and  d  =  —  4:. 

2.  Find  the  value  of  (a  —  x)(a  —  ^y){a  —  ^z). 

3.  If  x^  +  7x  +  c  is  exactly  divisible  by  x  +  4,  what  is  the 
value  of  c  ? 

4.  Divide  x^  —  y^  —  z'^ +2yz  by  a;  +  ?/  —  0,  and  find  the  co- 
efficient of  X  in  the  quotient  obtained  by  dividing  8  x*  +  xy^  —  y^ 
by  x-^y. 

5.  Show  that 

(g2+  62)  (c2+  c?2)  =  (ac  +  6(^)2+  (gd!  -  6c)2=  (ac  -  bdy-i-  (ad  +  6c)2. 

6.  Show  that 

(6  -  c)2  +  (a  -  6)  (g  -  c)  =  (c  -  a)2  +  (6  -  c)  (6  -  g) 
=  (g  -  6)2 +  (c-g)(c- 6). 

G.  1.   Simplify  5 x  -  3[2  a;  +  9  ?/  -  2{3  a:  -  4(?/  -  x)}]. 

2.  Subtract  6  g2  -  4  g6  +  7  62  from  2  g2  -  4  g6  +  2  62,  and  from 
the  remainder  subtract  4  g2  _  6  g6  -}-  5  62. 

3.  Multiply  g2  -1-  62  +  1  -  g6  -  g  -  6  by  g  +  6  +  1. 

4.  Divide  g^  -)-  6^  -  8  c^  +  3  g26  +  3  g62  by  g  +  6  -  2  c. 


MISCELLANEOUS   EXAMPLES   I.  91 

5.  The  product  of  two  algebraical  expressions  is  x^  —  64  x,  and 
one  of  the  expressions  is  x^  -  2  x  +  4  ;  what  is  the  other  ? 

6.  Prove  that 

(1  +  X+  X2)(l  _  X  +  X2)(l  -  X2  +  X*)(l  _  X*  +  X8)  =  1  +  X8  +  Xl6, 

and  that 

a3(6-c)  +  63(c_a)  +  c3(a-5)  +  (6-c)(c-a)(a-6)(a  +  ft  +  c)  =  0. 


92  SIMPLE  EQUATIONS. 


CHAPTER  VI. 

Simple  Equations. 

90.  A  statement  of  the  equality  of  two  algebraical 
expressions  is  called  an  equation,  and  the  two  equal  ex- 
pressions are  called  the  members  or  sides  of  the  equation. 

When  the  statement  of  equality  is  true  for  all  values 
of  the  letters  involved,  the  equation  is  sometimes  called 
an  identical  equation.  An  identical  equation  is,  however, 
generally  called  an  identity,  and  the  name  equation  is 
reserved  for  those  cases  in  which  the  equality  is  only 
true  for  certain  particular  values  of  the  letters  involved. 

Thus  a  +  «  =  2  a,  and  (a  +  hy  =  a^  -\-2ah  +  b^,  which  are  true 
for  all  values  of  a  and  6,  are  identities ;  and  5  a  +  2  =  12,  which 
is  only  true  when  a  is  2,  is  an  equation. 

In  an  identity  the  sign  =  is  frequently  used  instead  of 
the  sign  = .     Thus,  a  +  a  =  2a. 

Note.  —  Eor  the  sake  of  distinction,  a  quantity  which  is  sup- 
posed to  be  known,  but  which  is  not  expressed  by  any  particular 
arithmetical  number,  is  represented  by  one  of  the  first  letters  of 
the  alphabet,  a,  b,  c,  etc.,  and  a  quantity  which  is  unknown,  and 
which  is  to  be  found,  is  represented  by  one  of  the  last  letters  of  the 
alphabet,  x,  y,  or  z. 

91.  To  solve  an  equation  is  to  find  the  value,  or  values, 
of  the  unknown  quantity  for  which  the  equation  is  true ; 
and  these  values  of  the  unknown  quantity  are  said  to 


\ 


\  SIMPLE  EQUATIONS.  93 

satisfy  the   equation,   and  are   called  the  roots    of   the 
equation. 

Def.  In  more  precise  terms  we  therefore  define  a  root 
of  an  equation  to  be  any  quantity  which,  when  substi 
tuted  for  the  unknown  quantity,  reduces  the  equation 
to  an  identity. 

This  definition  supplies  us  with  a  convenient  test  for  a 
solution.     Thus  2  is  a  root  of 

because  when  2  is  substituted  for  x,  the   equation  is 
reduced  to  the  identity 

8-8  +  2  =  2. 

Ex.  1.   Show  that  1  is  a  root  of  each  of  the  following  equations : 

(i.)   x^-x  =  0,  (iv.)   2x8  +  x  =  3, 

(ii.)  rK2-3ic  +  2  =  0,  (v.)   aic^  -  ax  +  x  =  1, 

(iii.)   x8-l  =  0,  (vi.)   x^-lr^O. 

Ex.  2.   Show  that  o  is  a  root  of  each  of  the  following  equations : 
(i.)   x2_fl(2  =  o,  (iii.)  x«-a»  =  0, 

(ii.)  ax  —  a^  +  x  =  a^  (iv.)   x*  —  ax^  +  x*  _  ^x, 

(v.)  x2  -  (a  +  9)x  +  a6  =  0. 

92.  An  equation  which  contains  only  one  unknown 
quantity,  x  suppose,  is  said  to  be  of  the  first  degree  when 
X  occurs  only  in  the  first  power ;  it  is  said  to  be  of  the 
second  degree  when  a^  is  the  highest  power  of  x  which 
occurs,  and  so  on. 

Equations  of  the  first  degree  are,  however,  generally 
called  simple  equations,  and  equations  of  the  second  degree 
are  generally  called  quadratic  equations. 


94  SIMPLE   EQUATIONS. 

In  the  present  chapter  we  shall  only  consider  simple 
equations  which  contain  one  unknown  quantity. 

93.  In  the  solution  of  equations  frequent  use  is  made 
of  the  following  axioms : 

(i.)  If  we  add  to  equals  the  same  quantity,  or  equal 
quantities,  the  sums  will  be  equal. 

Thus,  ii  a  =  b,  then  a  +  c  =  b  +  c. 

(ii.)  If  we  take  from  equals  the  same  quantity,  or 
equal  quantities,  the  differences  will  be  equal. 

Thus,  if  a  =  6,  then  a  —  c  =  b  —  c. 

(iii.)  If  we  multiply  equals  by  the  same  quantity,  or 
by  equal  quantities,  the  products  will  be  equal. 

Thus,  if  a  =  6,  then  ac  =  be. 

(iv.)  If  we  divide  equals  by  the  same  quantity  not 
zero,  or  by  equal  quantities,  the  quotients  will  be  equal. 

Thus,  if  a  =  6,  then  a  ^  c  =  b  -i-  c.  But  although  2x0  =  3x0, 
we  cannot  divide  by  0  and  say  that  2  =  3. 

94.  Let  a  -f-  b  =  c  —d. 

Add  —  6  to  both  sides,  then  the  equality  still  holds 
good  by  Axiom  (ii.)  > 

.-.  a-{-b  —  b  =  c  —  d  —  bf 
that  is,  a  =  c  —  d  —  b. 

Thus  the  term  b  has  been  cancelled  from  one  side  of 
the  equation,  and  it  appears  on  the  other  side  with  its 
sign  changed  from  +  to  — . 


SIMPLE  EQUATIONS.  95 

Again,  add  d  to  both  sides  of  the  last  equation  j  then, 
by  Axiom  (i.), 

a-{-d  =  c  —  d  —  b-{-d'y 

.'.  a  -\-  d  =  c  —  b. 

Thus  the  term  d  has  been  cancelled  from  one  side  of 
the  equation,  and  it  appears  on  the  other  side  with  its 
sign  changed  from  —  to  -|-. 

We  can  proceed  in  a  similar  manner  in  any  other 
case ;  hence  any  term  may  be  moved  from  one  side  of  an 
equation  to  the  other,  provided  its  sign  is  changed. 

When  terms  are  changed  from  one  side- of  an  equation 
to  the  other  side,  they  are  said  to  be  transposed. 

95.  We  may  change  the  signs  of  all  the  terms  of  an 
equation;  for,  by  Axiom  (iii.),  we  do  not  destroy  the 
equality  by  multiplying  both  sides,  and  therefore  every 
term,  by  —1;  and  this  multiplication  will  change  the 
sign  of  every  term. 

96.  The  method  of  solving  simple  equations  will  be 
seen  from  the  following  examples : 

Ex.  1.   Solve  8x  +  7  =  4a;+27. 

Transpose  the  terms  4  x  and  7,  then 

8a; -4a:  =  27 -7. 
Combine  like  terms,  then 

4x  =  20. 
Divide  both  sides  by  4,  the  coefl&cient  of  x ;  then 
x  =  6. 


96  SIMPLE  EQUATIONS. 


Ex.  2.    Solve 

6a;-7: 

=  7a;-15. 

By  transposition,  we  have 

6X  —  1X: 

=  -  15  +  7. 

Combining  like  terms, 

,  we 

get 

-2X: 

=  -8. 

Divide  both  sides  by  - 

-2, 

the  coefficient  of  x ;  then 

• 

X: 

=  4. 

Ex.  3.  Solve 

|  +  - 

4^2 

Multiply  every  term  by  4,  the  L.  C.  M.  of  the  denominators,  then 
the  fractions  will  be  got  rid  of,  and  we  shall  have 

2  X  +  8  =  ic  +  10. 

Transposing,  2aj  —  x=10  —  8; 

,:x  =  2. 

Ex.  4.  Solve  ^-t^  _  1  (a;  _  1)  z=  1. 

Multiply  by  12  to  get  rid  of  fractions  ;  then 

\H^  +  i)-¥-(^-i)=i2, 

that  is,  3  («  +  1)  -  4  (a;  -  1)  =  12  ; 

.-.  3ic  +  3-4x  +  4  =  12. 
Transposing,  3aj  —  4a;  =  12  —  3— 4; 

r.  —  x  =  6, 
or,  by  changing  the  signs,  x  =  —  5. 

Ex.  5.  Solve  ax -}-  b^  =  bx -[-  a^. 

By  transposition,  ax—  bx  =  a^  —  b^, 

that  is,  (a  -  &)  X  =  a2  -  bK 

Divide  by  a  —  6,  the  coefficient  of  x ;  then 

X  =  (a2  -  62)  -^  (a  -  6)  =  a  +  6. 


SIMPLE  EQUATIONS.  97 

Note.  —  The  student  should  test  his  results  by  seeing  that  the 
values  obtained  really  satisfy  the  given  equations,  that  is,  reduce 
them  to  identities. 

For  example,  if  we  put  5  for  x  in  Ex.  1,  we  have 

8x5  +  7  =  4x5  + 27, 

that  is,  40  +  7  =  20  +  27, 

which  is  clearly  true. 

Again,  if  we  put  a  +  6  for  x  in  Ex.  5,  we  have 

a  (a  +  6)+  62  =  6  (a  +  6)+  a\ 

that  is,  a^j^  ah  +  b'^  =  ab  +  b^-\-  a% 

which  is  an  identity.     [See  Art.  91.] 

97.  From  the  above  it  will  be  seen  that  the  different 
steps  in  the  process  of  solving  a  simple  equation  are  as 
follows.  First,  clear  the  equation  of  fractions,  and  per- 
form all  the  algebraical  operations  which  are  indicated. 
Then  transpose  all  the  terms  into  which  the  unknown 
quantity  enters  to  one  side  of  the  equation,  and  all  the 
other  terms  to  the  other  side.  Next  combine  all  the 
terms  which  contain  the  unknown  quantity  into  one 
term,  and  divide  by  the  coefficient  of  the  unknown  quan- 
tity ;  this  gives  the  required  root. 

Note.  —  In  order  to  acquire  a  habit  of  clear  and  accurate  thought, 
the  way  in  which  each  new  equation  is  derived  from  the  preceding 
should  always  be  indicated. 

A  beginner  should  be  cautioned  against  the  common  mistake  of 
putting  a  meaningless  sign  of  equality  at  the  commencement  of  a 
line. 

98.  The  following  are  additional  examples  of  simple 
equations. 


98  SIMPLE   EQUATIONS. 

Ex.  1.  Solve  (x  -  1)  (a;  -  2)  +  5  =  (x  +  l)^. 

Removing  the  brackets,  we  have 

a;2 -3x4-2  + 5  =  a;2 +  2x4-1. 
Transposing,  we  have 

x2-3x-x2-2x  =  l-2-5; 
.-.   -5x=  -6. 
Divide  by  —  5  ;  then  ^  =  f  • 

Ex.  2.    Solve  3  (x  -  1)  -  {3  X  -  (2  -  x)}  =  5. 
Removing  the  brackets,  we  have 

3x-3-3x  +  2-x  =  5; 

.-.  3x-3x-x  =  5  +  3-2; 
.♦.  -X  =6, 
or  ^  X  =  —  6. 

Ex.   3.    Solve3x2-l=(3x  +  2)(x-5). 
Removing  the  brackets,  we  have 

3x2-1  =  3x2- 13x- 10. 
Transposing,  we  have 

3x2-3x2  +  13x=  -10  +  1. 
Hence  .  13x  =  —  9 ; 

.-.  x=-tV 

Ex.  4.    Solve  a(x  -  a)  +  6(x  -  &)  =  2  ah. 
Removing  the  brackets,  we  have 

ax  -  a2  _|_  5x  -  &2  =  2  ah. 
Transposing,  we  have 

ax  +  &x  =  a2  4.  52  _|.  2  ah, 
that  is,  x(a  +  6)  =  (a  +  hY  ; 

.-.  a;=(a  + &)2-(a  +  &)=a+ 6. 


SIMPLE   EQUATIONS. 


99 


EXAMPLES  XXI 
Solve  the  following  equations  : 

1.  3a;  +  4  =  ic  +  10. 

2.  x  +  7  =  4a;  +  4. 

3.  5ic-12=6x-8. 

4.  7a;  +  19  =  5x  +  7. 

5.  3(x-2)=2(x-3). 


14.    x  + 


6.  5(x  +  2)=3(x+3)-fl. 

7.  a;-(4-2x)=7(x-l). 

8.  5(4-3x)=7(3-4x). 

9.  2(x-3)=5(x+l)  +  2x-l. 
10.   4(l-x)+3(2  +  x)=13. 

11.  2(x  -  2)  +  3(x  -  3)  +  4(x  -  4)  -  20  =  0. 

12.  2(x-l)-3(x-2)+4(x-3)+2=0. 

13.  5x  +  6(x+ l)-7(x  +  2)-8(x+ 3)  =  ©. 

2 


10. 


16. ^  +  ^^ 

2  2 


3. 


18.  K2-a;)-K5^  +  21)  =  x  +  3. 

19.  ^(x-2)+KaJ-3)+Ka^-4)=10. 


^ 


22.   ^+  1  I  a?  +  2  ,  g;H-3 


23.  2x-[3-{4x  +  (x-l)}-5]=8. 

24.  l-2{x-3(l  +  x)}  =  0. 

25.  (x+l)(x  +  2)  =  (x-2)(x-4). 

26.  (x-l)(x-2)  =  (x-3)(x-4). 

27.  2x2=(x  +  l)2  +  (a;  +  3)2. 

28.  3x2=(x+l)2-f(x  +  2)2  +  (x  +  3)2. 


100  SIMPLE  EQUATIONS. 

29.  {x-2){x  -b)  +  (x-  3)(x-  4)=  2(x  -  4)(a;  -  5). 

30.  (x  -  1)2  +  ^(x  -  3)2  =  b{x  +  5)2. 

31.  5(x  +  l)2  +  7(a;  +  3)2  =  12(x  +  2)2. 

32.  (a;-l)(x-4)=2a;  +  (x-2)(a;-3). 

33.  (X  -  1)3  +  (X  -  2)3  +  (x-  3)3  =  3(x  -  l)(x  -  2)(x  -  3). 

34.  ^+i_2^:zi+ii^0.        35.    ^±i_21_4^^ 

2  5  *  8  2 

36.  .5x +  37.5  =  5.25  a; -1.  37.    .25x  +  4  -  .375x=  .2x  -  9. 

38.  .15  a;  +  1.2  -  .875  X  +  .375  =  .0625  a;. 

39.  1.2x-K.18ic-.05)=.4x+8.9. 

40.  a{x-  a)=h(x-h).  41.   2(x  -  a)+ 3(x  -  2a)  =  2a. 

42.  \(x+  a  +  h)-\-l{x+  a-  b)  =  b. 

43.  (a  +  b)x  +  (a  -  b)x  =  a^.      44.    (a  +  6)x +  (&  -  a)x  =  62, 

45.  i(a  +  x)+ i(2«  + x)+ K3a  +  x)=  3  a. 

46.  ^  +  ^  =  a2^_52.  47.    (a+6x)(6+ax)=a6(x2-l). 

6        a 

48.  (a2  +  x)  (62  +  x)  =  (a6  +  x)2. 

49.  a(x  +  a)+ 6(6- x)  =  2  a6. 

60.    (x  +  a  +  6)2+  (X  +  a  -  6)2  =  2 x2. 

51.    (x-a)(x-6)  +  (a  +  6)2=(x  +  a)(x+ 6)7) 

62.    (x  +  a  +  6  +  c)  (x  +  a  -  6  -  c)  = 

(x  -  a  —  6  +  c)  (x  —  a  +  6  —  c). 

53.    ax(x  +  a)  +  6x(x  +  6)  =  (a  +  6) (x  +  a) (x  +  6). 

64.    (X  -  a)3  +(x  -  6)3  +  (x  -  c)3  =  3(x  -  a)(x  ~  6)(x  -  c). 


PKOBLEMS.  101 


CHAPTER  VII. 
Problems. 

99.  With  some  of  the  general  methods  of  algebra 
now  at  our  command,  we  return  to  the  subject  introduced 
in  the  first  chapter ;  namely,  the  solution  of  problems. 

In  order  to  solve  a  problem,  the  relations  between  the 
known  and  unknown  quantities  must  be  expressed  by- 
means  of  algebraical  symbols :  we  thus  obtain  equations, 
the  roots  of  which  are  the  required  values  of  the  unknown 
quantities. 

100.  In  the  present  chapter  we  shall  only  consider 
problems  in  which  there  is  one  unknown  quantity,  and 
in  which  the  relation  between  the  known  and  the  un- 
known quantities  is  expressed  algebraically  by  means  of 
a  simple  equation. 

Of  such  problems  the  following  are  examples : 

Ex.  1.  A  has  $  20,  and  B  has  $  3.75.  How  much  must  A  give 
to  B  in  order  that  he  may  have  just  four  times  as  much  as  B  ? 

Let  X  be  the  number  of  dollars  that  A  gives  to  B. 

Then  A  will  have  20  —  x  dollars,  and  B  will  have  3.75  +  x  dol- 
lars. But  A  now  has  four  times  as  much  as  B.  Hence  we  have 
the  equation 

20-a;  =  4(3.75-hx). 

^That  is  20-  x  =  lb  +  ix. 

Transposing         — x  —  4x  =  15  —  20, 
that  is  —  5  X  =  —  5. 


102  PROBLEMS. 

Divide  by  —  5,  then  x  =  l. 

Thus  A  must  give  one  dollar  to  B. 

Note.  —  It  should  be  remembered  that  x  must  always  stand  for  a 
number.  It  is  also  to  be  noticed  that  in  any  problem  all  concrete 
quantities  of  the  same  kind  must  be  expressed  in  terms  of  the 
same  unit ;  for  example,  in  the  above  all  sums  of  money  were 
expressed  as  dollars. 

Ex.  2.  A  man  has  12  coins,  some  of  which  are  half-dollars  and 
the  rest  dimes,  and  the  coins  are  worth  4  dollars  altogether.  How 
many  are  there  of  each  kind  ? 

Let  X  be  the  number  of  half-dollars  ;  then  12  —  ic  will  be  the 
number  of  dimes.  The  half-dollars  are  worth  ^  x  dollars,  and  the 
dimes  are  worth  xV(12  —  x)  dollars.  Hence,  since  the  coins  are 
worth  4  dollars  altogether,  we  have  the  equation 

4. 


Ix  +  jW2-x) 

Therefore 

lx-j\x 

at  is 

t\^ 

.:  IK  =  28  -  4 

7. 
Hence  there  are  7  half-dollars  and  5  dimes. 

Ex.  3.  A  father  is  six  times  as  old  as  his  son,  and  in  four  years 
he  will  be  four  times  as  old.     How  old  is  each? 

Let  the  son  be  x  years  old.  Then  the  father  must  he  6x  years 
old.  After  four  years  the  son  will  be  x  -f  4,  and  the  father  will  be 
6  oj  -I-  4  years  old.     Hence  by  the  question 


6x-{-4:  =  4(X-\-4:) 

that  is 

6a; +  4  =  4x4- 16. 

Hence 

6x-4x  =  16-4, 

that  is 

2  X  =  12 ; 

.-.  x  =  6. 

Hence  the  son  is  6  years  old,  and  the  father  is  36  years  old. 


PROBLEMS.  103 

Ex.  4.  A  can  do  a  piece  of  work  in  12  hours  which  B  can  do 
in  4  hours.  A  begins  the  work,  but  after  a  time  B  takes  his  place, 
and  the  whole  work  is  finished  in  6  hours  from  the  beginning. 
How  long  did  A  work  ? 

Let  X  —  the  number  of  hours  that  A  worked. 
Then  6  —  a;  =  the  number  of  hours  that  B  worked. 
Since  A  can  do  the  whole  work  in  12  hours,  the  part  done  by  A 
in  1  hour  is  ^^. 

Therefore  the  part  done  by  A  altogether  is  — • 

Since  B  can  do  the  whole  work  in  4  hours,  the  part  done  by  B 
in  1  hour  is  \. 

Therefore  the  part  done  by  B  altogether  is  |(6  —  x). 

But  A  and  B  together  do  the  whole  of  the  work.  Hence  we 
have  the  equation 

Multiply  by  12,  then 

a;  +  3(6-x)=12, 
that  is  a;+ 18 -3ic  =  12. 

Transposing  —  2  x  =  —  6 ; 

.-.  a;  =  3. 
Hence  A  worked  for  3  hours. 

Ex.  6.  Find  the  time  between  3  and  4  at  which  the  hands  of 
a  clock  are  together. 

Suppose  that  the  hands  are  together  at  x  minutes  after  3  o'clock. 

At  3  o'clock  the  hour-hand  is  15  minute-spaces  in  front  of  the 
minute-hand,  and  after  x  minutes  they  are  together.  Hence 
while  the  minute-hand  moves  through  x  minute-spaces  the  hour- 
hand  will  move  through  x  —  Vo  such  spaces.  But  the  minute- 
hand  moves  twelve  times  as  fast  as  the  hour-hand,  and  therefore 
in  any  time  the  minute-hand  passes  over  twelve  times  as  many 
minute-spaces  as  the  hour-hand. 


104  PROBLEMS. 

Hence  jc  =  12(ic  -  15), 

that  is  cc  =  12  a;  -  180  ; 

.-.  llx  =  180, 
or  x=  Vt-  =  16/t. 

Thus  the  time  required  is  IQfj  minutes  past  3  o'clock. 

EXAMPLES    XXII. 

1.  Find  two  numbers  whose  sum  is  200,  and  whose  difference 

is  2.      ' ;'  . 

2.  r^id  two  numbers  whose  sum  is  66,  and  whose  difference 
is  20.  {S 

3.  Divide  25  into  two  parts  whose  difference  is  5.    /  ^ 

4.  Divide  100  into  two  parts  whose  difference  is  45.  ^^  ->< 

5.  What  number  is  that  to  which  if  you  add  40  the  sum  will 
be  three  times  the  original  number  ?  "a^  t? 

6.  What  number  is  that  from  which  if  you  take  away  14  the 
remainder  is  one-third  of  the  original  number  ?  v  f 

7.  The  difference  of  two  numbers  is  15,  and  one  number  is 
four  times  the  other.     Find  the  numbers.      • 

8.  Find  two  numbers  whose  difference  is  10,  and  one  of  which 
is  three  times  the  other. 

9.  The  sum  of  two  numbers  is  38,  and  one  of  them  exceeds 
twice  the  other  by  2.     What  are  the  numbers  ?     '    >. 

10.  Find  two  numbers  the  sum  of  which  is  31,  and  which  are 
such  that  one  of  them  is  less  by  2  than  half  the  other. 

11.  Find  a  number  whose  fourth  part  exceeds  its  fifth  part  by  2. 

12.  Find  a  number  whose  third  part  exceeds  its  seventh  part 
by  80. 

13.  Find  a  number  which  when  multiplied  by  4  is  as  much 
above  35  as  it  was  originally  below  it. 


PROBLEMS.  105 

14.  Find  a  number  which  when  multiplied  by  8  exceeds  27  as 
much  as  27  exceeds  the  original  number. 

15.  Four  times  the  difference  between  the  fourth  and  fifth  parts 
of  a  certain  number  exceeds  by  4  the  difference  between  the  third 
and  seventh  parts.     What  is  the  number  ? 

16.  Fifty  times  the  difference  between  the  seventh  and  eighth 
parts  of  a  certain  number  exceeds  half  the  number  by  44.  What 
is  the  number  ? 

17.  Divide  100  into  two  parts  such  that  three  times  one  of  the 
parts  plus  five  times  the  other  is  410. 

18.  Divide  100  into  two  parts  such  that  twice  one  part  is  equal 
to  three  times  the  other. 

19.  The  difference  of  two  numbers  is  20,  and  one-half  of  one 
of  the  numbers  is  equal  to  one-fifth  of  the  other.     Find  them. 

20.  The  sum  of  two  numbers  is  36,  and  their  difference  is  half 
the  greater.     Find  them. 

21.  A  has  $  100,  and  B  has  §20  ;  how  much  must  A  give  B  in 
order  that  B  may  have  half  as  much  as  A  ? 

22.  A  and  B  play  for  a  stake  of  5s.  If  A  loses,  he  will  have  as 
much  as  B,  but  if  A  wins,  he  will  have  three  times  as  much  as  B. 
How  much  has  each  ? 

23.  A  and  B  have  $  50  between  them.  A  wins  from  B  as  much 
as  he  had  originally,  and  he  then  has  three  times  as  much  as  B. 
What  had  A  at  first  ? 

24.  A,  B,  and  C  have  a  certain  sum  between  them.  A  has  one- 
half  of  the  whole,  B  has  one-third  of  the  whole,  and  C  has  $  50. 
liow  much  have  A  and  B  ? 

25.  A  and  B  together  have  $75,  and  A  has  $5  more  than  B  ; 
how  much  has  each  ? 

26.  A  has  $  5  more  than  B,  B  has  .f  20  more  than  C,  and  they 
have  1 360  between  them.     How  much  has  each  ? 


106  PROBLEMS. 

27.  A  has  ^  15  more  than  B,  B  has  $  5  less  than  C,  and  they 
have  $65  between  them.     How  much  has  each  ? 

28.  A  has  $  5  less  than  B,  C  has  as  much  as  A  and  B  together, 
and  they  have  3 100  between  them.     How  much  has  each  ? 

29.  Divide  $150  among  10  men,  20  women,  and  40  children, 
giving  to  each  man  $3.75  more  than  to  each  child,  and  to  each 
woman  as  much  as  to  two  children. 

30.  Divide  $  15  among  3  men,  5  women,  and  20  children,  giving 
to  each  man  one  dollar  more  than  to  each  woman,  and  to  each 
child  half  as  much  as  to  each  woman. 

31.  A  man  of  40  has  a  son  10  years  old ;  in  how  many  years 
will  the  father  be  three  times  as  old  as  the  son  ? 

32.  One  man  is  70  and  another  is  50  years  of  age  ;  when  was  the 
first  twice  as  old  as  the  second  ? 

33.  A  father's  age  is  three  times  that  of  his  son,  and  in  10  years 
it  will  be  twice  as  great ;  how  old  are  they  ? 

34.  In  5  years  a  father  will  be  just  four  times  as  old  as  his 
daughter,  and  in  10  years  he  will  be  just  three  times  as  old  ;  how 
old  is  he  now  ? 

35.  A  sum  of  money  is  divided  between  three  persons  —  A,  B, 
and  C  —  in  such  a  way  that  A  and  B  have  $  60  between  them,  A 
and  C  have  $65,  and  B  and  C  have  $  75.     How  much  has  each  ? 

36.  Four  persons  —  A,  B,  C,  D  —  are  possessed  of  certain  sums 
of  money,  such  that  A  and  B  together  have  $  49,  A  and  C  together 
have  $  51,  B  and  C  together  have  $  53,  and  A  and  D  together  have 
$  47.     How  much  has  each  ? 

37.  What  is  the  price  of  beef  if  a  reduction  of  20  per  «ent  in 
the  price  would  enable  a  purchaser  to  obtain  6  lbs.  more  for  five 
dollars  ? 

38.  How  much  are  eggs  a  score,  when  a  rise  of  20  per  cent  in 
the  price  would  make  a  difference  of  80  in  the  number  which  could 
be  bought  for  five  dollars  ? 


PROBLEMS.  107 

39.  A  man  leaves  one-half  of  his  property  to  his  wife,  one-third 
to  a  son,  and  the  remainder  (which  is  $  2000)  to  a  daughter.  How 
much  did  he  leave  altogether  ? 

40.  A  man  left  his  property  to  be  divided  among  his  three 
children  in  such  a  way  that  the  share  of  the  eldest  was  to  be  twice 
that  of  the  second,  and  the  share  of  the  second  twice  that  of  the 
youngest.  It  was  found  that  the  eldest  received  $  750  more  than 
the  youngest.     How  much  did  each  receive  ? 

41.  A  purse  contains  30  coins  which  altogether  amount  to 
$  136,50.  A  certain  number  of  the  coins  are  dimes,  one-fifth  of 
that  number  are  half-eagles,  and  the  rest  are  eagles.  Find  the 
number  of  each. 

42.  A  purse  contains  36  coins  which  altogether  amount  to 
$  52.80.  A  certain  number  of  the  coins  are  eagles,  there  are  three 
times  as  many  half-eagles,  and  the  rest  are  dimes.  Find  the  num- 
ber of  each. 

43.  Find  the  time  between  6  o'clock  and  6  o'clock  when  the 
hands  of  a  watch  are  together. 

44.  Find  at  what  times  between  9  and  10  o'clock  the  hands  of  a 
watch  are  at  right  angles  to  one  another. 

45.  There  are  two  numbers  one  of  which  exceeds  the  other  by 
3,  while  its  square  exceeds  the  square  of  the  other  by  99.  Find 
the  numbers. 

46.  In  a  mixture  of  spirits  and  water  half  of  the  whole  plus 
25  gallons  was  spirit ;  and  a  third  of  the  whole  minus  5  gallons 
was  water  ;  how  many  gallons  were  there  of  each  ? 

47.  A  garrison  of  1000  men  having  provisions  for  60  days  was 
reinforced  after  10  days,  and  from  that  time  the  provisions  only 
lasted  20  days.    Find  the  number  in  the  reinforcement. 

48.  A  laborer  was  engaged  for  36  days,  upon  the  condition 
that  he  should  receive  $  3.25  for  every  day  he  worked,  but  should 
pay  $0.75  for  every  day  he  was  idle.  At  the  end  of  the  time  he 
received  $  29.     How  many  days  did  he  work  ? 


108  PROBLEMS. 

49.  A  sum  of  money  is  divided  among  three  persons.  The 
first  receives  $  10  more  than  a  third  of  the  whole  sum  ;  the  second 
receives  $  15  more  than  a  half  of  what  remains  ;  and  the  third 
receives  what  is  over,  which  is  $  70.     Find  the  original  sum. 

50.  A  purse  of  sovereigns  is  divided  amongst  three  persons,  the 
first  receiving  half  of  them  and  one  more,  the  second  half  of  the 
remainder  and  one  more,  and  the  third  6.  Find  the  number  of 
sovereigns  the  purse  contained. 

51.  From  a  sum  of  money  $  20  more  than  its  half  was  taken 
away ;  from  the  remainder  $  30  more  than  its  third  part ;  and 
from  what  then  remained  f  40  more  than  its  fifth  part ;  after  which 
there  was  nothing  left.     What  was  the  sum  ? 

52.  At  a  cricket  match  the  caterer  provided  dinner  for  24 
persons,  and  fixed  the  price  so  as  to  gain  121  per  cent  upon  his 
outlay.  Three  of  the  cricketers  were  absent.  The  remaining  21 
paid  the  fixed  price  for  their  dinner,  and  the  caterer  lost  25 
cents.     What  was  the  charge  for  dinner  ? 

53.  A  can  do  a  piece  of  work  in  30  days,  which  B  can  do  in 
20  days.  A  begins  the  work,  but  after  a  time  B  takes  his  place, 
and  the  whole  work  is  finished  in  25  days  from  the  beginning. 
How  long  did  A  work  ? 

54.  A  can  do  a  piece  of  work  in  20  days,  which  B  can  do  in 
30  days.  A  begins  the  work,  but  after  a  time  B  takes  his  place 
and  finishes  it ;  and  B  worked  for  10  days  longer  than  A.  How 
long  did  A  work  ? 

55.  How  many  men  are  there  in  a  regiment  which  can  be  drawn 
up  in  two  hollow  squares  with  the  men  3  and  5  deep  respectively, 
if  the  one  square  will  just  fit  within  the  other,  and  the  same  num- 
ber of  men  be  in  each  square  ? 


SIMULTANEOUS  EQUATIONS.  109 


CHAPTER  VIII. 

Simultaneous  Equations  of  the  First  Degree. 

101.  A  single  equation  which  contains  two  unknown 
quantities  can  be  satisfied  by  an  indefinite  number  of 
pairs  of  values  of  the  unknown  quantities.  Consider, 
for  example,  the  equation  a;  — 3?/ =  24.  It  is  clear  that 
the  equation  is  satisfied  by  the  values  y  =  0  and  re  =  24; 
or  2/  =  1  and  a;  =  27 ;  or  y  =  2  and  x  =  30.  In  fact,  we 
may  suppose  y  to  be  equal  to  any  quantity  k,  provided 
that  X  is  taken  equal  to  24  +  3  k. 

If  there  are  two  equations  containing  two  unknown 
quantities,  each  equation  taken  by  itself  can,  as  we  have 
just  seen,  be  satisfied  in  an  indefinite  number  of  ways ; 
but  this  is  not  the  case  when  both  equations  are  to  be 
satisfied  by  the  same  values  of  the  unknown  quantities. 

Def.  Two  or  more  equations  which  are  satisfied  by  the 
same  values  of  the  unknown  quantities  contained  in 
them,  are  called  simultaneous  equations. 

102.  The  degree  of  an  equation  which  contains  the  two 
unknown  quantities  x  and  y,  is  the  degree  of  that  term 
which  is  of  the  highest  dimensions  in  x  and  y.     [Art.  67.] 

Similarly  the  degree  of  an  equation  which  contains 
the  unknown  quantities  «,  y,  and  z  is  the  degree  of  that 
term  which  is  of  the  highest  dimensions  in  x,  y,  and  z. 


110  SIMULTANEOUS   EQUATIONS. 

Thus  4a;  +  5?/ =  20  is  an  equation  of  the  Jirst  degree; 
and  Soc^  —  2xy  =  7  is  an  equation  of  the  second  degree. 

The  equations  ax^  -|-  6?/^  +  c  =  0  and  x-\-  y  =  xy  are  also 
of  the  second  degree. 

103.  In  the  present  chapter  we  shall  show  how  to 
solve  simultaneous  equations  of  the  first  degree. 

It  will  be  seen  that  any  two  simultaneous  equations 
containing  two  unknown  quantities  can  be  solved  by  de- 
ducing from  the  given  equations  a  third  equation  from 
which  one  of  the  unknown  quantities  is  absent ;  and 
this  will  give  the  value  of  that  unknown  quantity  which 
is  retained. 

104.  Elimination  by  Addition  and  Subtraction.  We  will 
show  how  to  solve  two  simultaneous  equations  of  the 
first  degree  by  taking  as  an  example  the  following  pair 
of  equations : 

3a; +  52/ =  22, 
and  7a;  — 42/ =  20. 

Multiply  each  member  of  the  first  equation  by  7,  and 

each  member  of  the  second  equation  by  3 ;  the  equations 

then  become 

21a; +  352/ =  154, 

and  21a; -122/=    00. 

The  coefficients  of  x  are  now  the  same  in  the  two  equa- 
tions ;  hence  if  we  subtract  each  member  of  the  second 
equation  from  the  corresponding  member  of  the  first,  we 
shall  obtain  an  equation  from  which  x  is  absent :  the 

equation  will  be 

472/ =  94; 

whence  2/  =  2- 


SIMULTANEOUS   EQUATIONS.  Ill 

If  we  put  this  value  of  y  in  the  first  of  the  given 
equations,  we  have  3  a; -1-10  =  22;  hence  3  a;  =  22  — 10, 
or  ic=4.  Observe  that  this  value  of  x  could  equally  well 
be  found  fi  m  the  second  equation. 

We  may,  if  we  please,  find  the  value  of  x  before  finding 
the  value  of  y.  To  do  this,  multiply  the  first  equation 
by  4,  and  the  second  by  5 ;  the  equations  then  become 

12  a; -h  20  2/ =  88, 

and  35a; -202/ =  100. 

The  coefficients  of  y  in  the  two  equations  are  now 
equal  in  magnitude  but  opposite  in  sign  ;  hence  if  we  add 
each  member  of  the  second  equation  to  the  correspond- 
ing member  of  the  first,  we  shall  obtain  an  equation 
from  which  y  is  absent ;  the  equation  will  be 

47  a;  =188; 
whence  a;  =  4. 

We  .can  now  find  the  value  of  y  by  putting  a;  =  4  in 
the  first  of  the  given  equations. 

It  will  be  seen  from  the  above  example  that  in  order 
to  solve  twc  simultaneous  equations  of  the  first  degree 
we  can  proceed  as  follows :  Multiply  the  given  equa- 
tions by  numbers  such  that  in  the  resulting  equations 
the  coefficients  of  one  of  the  unknown  quantities  may  be 
equal  in  magnitude  ;  then  by  addition  or  subtraction  we 
shall  obtain  a  simple  equation  which  contains  only  one 
of  the  unknown  quantities,  and  which  can  be  solved  as 
in  the  preceding  chapter. 

The  unknown  quantity  which  has  been  got  rid  of  by 
the  above  process  is  said  to  have  been  eliminated,  and  the 


112  SIMULTANEOUS   EQUATIONS. 

particular  method  above  explained  is  known  as  elimina- 
tion by  addition  and  subtraction. 

It  is  generally  best  to  eliminate  that  unknown  quan- 
tity which  has  the  smaller  coefficients  in  the  two  equar 
tions. 

105.   The  following  are  additional  examples : 
Ex.1.    Solve  3x  +  22/=  13,  7ic  + 3y  =  27. 
Multiply  the  equations  by  3  and  2  respectively  ;  we  then  have 
9a; +  6?/ =  39, 
and  14  jc  -f-  6  ?/  =  54. 

By  subtraction  we  eliminate  y,  and  have 
-6  a;  =  -15; 
.-.  x  =  3. 
Put  this  value  of  x  in  the  first  of  the  given  equations  ;  then 
9+2y  =  13; 
.-.  y  =  2. 
Thus  X  =  3,  y  =  2  are  the  values  required. 

Ex.2.    Solve2x~32/  + 14  =  0,  -4x-|- 52/  =  26. 
We  have  2  x  -  3  ?/  =  -  14, 

and  -4x  +  5?/  =  26. 

Multiply  the  first  equation  by  2  ;  then 

4x-62/  =  -28. 
From  the  last  two  equations  we  have  by  addition 

or  2/  =  2. 


SIMULTANEOUS  EQUATIONS.  113 

We  then  have  from  the  first  equation, 
2x-6=-14, 
that  is  2a:  =  -14  +  6=-8; 

.-.  x  =  -i. 
Hence  the  values  required  are 

X  =  —  4,  y  =  2. 

Ex.  3.  Solve  ^-^  =  9,    6x-^  =  29. 
3      4  4 

Clear  of  fractions  by  multiplying  the  equations  by  12  and  4 

respectively ;  then 

20a; -3y  =  108, 

24a;-7y  =  116. 

Multiply  the  first  equation  by  7  and  the  second  by  3 ;  then 

140x-21y  =  756, 

72  a;  -  21 2/  =  348. 

By  subtraction,  we  have 

68  a;  =  408; 

.-.  x  =  6. 

Then  120 -3?/ =  108, 

whence  «/  =  4. 

EXAMPLES  XXIIL 
Solve  the  equations : 

1.  7a; +  4 2/ =  1,  4.   8a;-21y  =  5,  7.   19a;  +  85y=350, 
9x  +  4y  =  3.                 6x+14y  =  -26.  17a;+119y=442. 

2.  3x  +  5?/  =  19,  5.   34x-152/=4,         8.    8x-lly  =  0, 
6x-4y  =  7.  51x  +  25?/  =  101.  25x-17y=139. 

3.  x-lly  =  l,  6.    39x-15y  =  93,        9.   3x-lly  =  0, 
llly-9x  =  99.  65x-|-17y=113.  19x-19y  =  8. 


114  SIMULTANEOUS   EQUATIONS. 

10.  ^+1=1,  12.    f+32/  +  14  =  0,    14.  ^±^  +  4?/  =  2, 

43  5^^^  11  2 

11.  f-^  =  l,  13.    ^+52/=:-4,         15.2^+^+^6=2, 
36      2  5  57 

^-^  =  ^.  l(4-5a.  =  4  2x-5y    a;+7^-. 

5       10      2  5  *  3  4' 

16.   4x-K2/-3)=5a:-3,  18.   ^^-^-^  =  0, 

2,+  i(2x-5)  =  21^^1I.  2x-5_ll-2y^Q^ 

6  5  7 

"•   |-J(y-2)-j(x-3)  =  0,  19.   «:=-?_2^  =  0, 

106.  Other  Methods  of  Elimination.  Instead  of  proceeding 
as  in  Art.  104,  we  may  solve  two  simultaneous  equations 
in  any  one  of  the  following  ways  : 

Ex.    Solve  the  equations  3  a;  —  5  y  =  2,  and  5  a;  —  2  y  =  16. 

(i.)  From  the  first  equation  we  have  ^x  =  5?/  +  2,  and  there- 
fore a;  =  1(5  y  4-  2).  Now  substitute  this  value  of  x  in  the  second 
equation,  and  we  get 

|(5y  +  2)-2^/  =  16, 
which  is  a  simple  equation  containing  only  one  of  the  unknown 
quantities  ;  the  solution  of  which  gives  y  =  2.    Then,  since  y  =  2, 

we  have 

x  =  KS^  +  2)=K10  +  2)=4. 

This  method  is  known  as  elimination  by  substitution. 

(ii.)  We  may  also  proceed  as  follows:  From  the  first  of  the 
given  equations  we  have  S  x  =  6  y  -{-  2,  or  a;  =  i  (5  ?/  +  2).  Also 
from  the  second  equation  we  have  5  x  =  2  ?/  +  16,  or  x  =  ^  (2  y  + 16) . 


SIMULTANEOUS   EQUATIONS.  115 

Hence,  by  equating  the  two  values  of  a;,  we  get    ♦ 

i(5y  +  2)=i(2y+16). 
The  last  equation  gives  y  —  2,  and  then,  as  before,  we 'find  that 

This  method  is  known  as  elimination  by  comparison. 

(iii.)  Or,  we  may  proceed  as  follows  :  Multiply  the  .first  of  the 
given  equations  by  A:,  an  arbitrary  multiplier,  and  to  the  resulting 
equation  add  the  second  of  the  given  equations.  Thus,  adding 
together 

3  A;x  -  5  A:?/  =  2  ^^  and  6  X  -  2  y  =  16, 

we  obtain  (3  A;  +  5)  x  -  (5  A;  +  2)  y  =  2  A;  4- 16. 

This  equation  is  true  for  all  values  of  k.  We  may  therefore  give 
to  k  two  values  in  succession  that  will  make,  first  the  coefficient  of 
x,  and  then  the  coefficient  of  y,  vanish.    Thus,  taking  A:  =  —  |,  we 

have  -  (  -  2j5  -I-  2)  y  =  -  ^  +  16, 

which  determines  y,  and  taking  A;  =  —  f ,  we  obtain 

(-|  +  6)a;=-  I  +16, 

which  determines  x.     This  method  is  known  as  elimination  by 
undetermined  (or  arbitrary)  multipliers. 

No  one  method  of  elimination  is  preferable  for  all  cases.  The 
practised  worker  in  algebra  chooses  that  form  which  is  best  suited 
to  each  example  as  it  arises. 

107.  The  following  are  examples  of  common  types  of 
simultaneous  equations : 

Ex.  1.  Solve  57  a;  +  25  2/  =  3772,  and  25  x  +  57  y  =  1148. 
We  have        67  x  +  25  ?/  =  3772, 
25x  +  57y  =  1148. 

In  this  example,  the  direct  application  of  the  method  of  Art.  104 
would  lead  to  troublesome  arithmetic  ;  but,  since  the  coefficients 
of  X  and  y  are  simply  interchanged  in  the  two  given  equations,  we 
should,  by  addition  or  subtraction,  obtain  equations  in  which  the 
coefficients  of  x  and  y  would  be  equal.     Hence  we  proceed  thus : 


116  SIMULTANEOUS   EQUATIONS. 

By  addition,  S2x-\-82y  =  4920  ; 

.♦.  cc  +  ?/ =  4920 -- 82  =  60. 
By  subtraction,  S2x-S2y  =  2624  ; 

.-.  X  -  ?/ =  2624  -  32  =  82. 

And  from  the  equations  x  +  y  =  QO,  x  —  y  =  S2,  we  obtain  at 
once  X  =  71,  and  y  =  —  11. 

Ex.2.   Solve  (X  -  l)(iy  -  2)-(x  -  2)  (y  -  1)=  -2, 

(X  -^2)(y  +  2)-(x-  2)  (y  -  2)=  32. 
On  removing  the  brackets,  the  first  equation  becomes 
xy-y-2x  +  2-xy  +  2y  +  x-2=-2, 
or  —x  +  y  =  —2..       (i.) 

The  second  equation  becomes 

xy  +  2x-\-2y  +  4:-xy  +  2x-}-2y-4:  =  S2, 
or  4a:  +  4?/  =  32     .     .      (ii.) 

Dividing  (ii.)  by  4,  we  have  x-{-  y  =  S       .     .     (iii.) 

From  (i.)  and  (iii.)  we  have  by  addition 

2  y  =  6,  and  therefore  y  =  S. 
From  (i.)  and  (iii.)  we  have  by  subtraction 
2  ic  =  10,  and  therefore  x  =  5. 

Ex.  3.   Solve  ^  +  -  =  8,     ^  +  ^  =  13. 
X     y  X     y 

These  equations  can  be  solved  as  two  simultaneous  equations 

containing  the  unknown  quantities  -  and  -  • 

X  y 

Multiply  the  equations  by  5  and  3  respectively  ;  then  we  have 

X       y 

and  1^  +  1^  =  39. 

X       y 


SIMULTANEOUS   EQUATIONS.  117 

Hence  we  have  by  subtraction 

2 

-  =  1,  and  therefore  y  =  2. 

y 

We  then  have  from  the  first  equation 

X     2 
that  is  '        §  =  8-2=6; 

X 

.'.  3  =  6a;,  or  x  =  ^. 
Thus  X  =  ^,  y  =  2  are  the  values  required. 

Ex.  4.   Solve  ax-{-  by  =  2  o.b,    bx  —  ay  =  b^  —  a^. 

Multiply  the  first  equation  by  a,  and  the  second  equation  by  b ; 
we  then  have 

a'^x  -h  aby  =  2  a^b, 
and  6%  —  aby  =  b^  —  a%. 

Hence  by  addition  a^x  +  6%  =  a^b  +  b^, 
that  is  (a^  +  b'^)x  =  b{a^  +  62); 

.'.  x  =  b. 
Put  this  value  for  x  in  the  first  of  the  given  equations ;  then 
ab  ■\-by  =  2ab\ 
.'.  by  =  ab; 
.-.  y  =  a. 
Thus  X  =  b,  y  =  a  B,re  the  required  values. 

Ex.5.   Solve6«H-3y -13  =  3a;-2y  +  7  =  2a;  +  6y-4. 

Since  6a;  +  3y-13  =  3a;-2?/  +  7, 
we  have  2x  +  5y  =  20 (i.) 

Also,  since  Bx-2y-^7  =  2x-^6y-4, 
we  have  x-Sy=  -11 (ii.) 

The  solution  of  (i.)  and  (ii.)  by  the  method  of  Art.  104  gives 
r  =  5  and  y  —  2. 


118  SIMULTANEOUS    EQUATION^. 

y  EXAMPLES  XXIV. 

Solve  the  equations : 

1.  17  X  +  23  ?/ =  86,  4.       29.x  +  85?/=  31, 
23  cc  +  17  ?/ =  74.  13ic-43y  =  95. 

2.  15ic+19y  =  18,  5.  x+.2?/  =  .3, 
19a:  +  15?/ =  50.                            1.7  a;  +  .01  y  =  .345. 

3.  51a;- 14?/ =  287,  6.  .5a;  +  ?/ =  2.75, 
14  a;  -  51 2/ =- 157.                      3.4  x  + -02?/ =  1.76. 

7.  3a;-4?/  +  2  =  5a;-6?/-2  =  7x  +  22/  +  4. 

8.  4a;-6?/-3  =  7a;  +  2?/-4=-2x  +  3?/  +  24. 

9.  3x  +  Z?/-2  =  ll?/-  — =  20. 

2i  5 

10.    5a;  +  ^-l  =  3?/+--2  =  4. 
5  3 

11    7  +  a;^9  +  y^ll  +  x4-y  ^g    ^-±1  =  l^-±-^  =  ^-+ 

3  5  7  '  '234 

13.  (a;+l)(?/  +  5)  =  (a;+5)(?/  +  l), 
a;2/  +  a;-f  y=(x  +  2)(?/  +  2). 

14.  a;2/-(a;-l)(?/-l)  =  6(?/-l),  a;-?/  =  l. 

15.  ?  _  §  ^  2,  19. 

X     y 

18  +  8  =  10. 

a;      ?/ 

J6.     -  +  -  =  7,  20. 

a;     ?/ 

§_?=ii. 

a;     2/ 

17.  §  +  ^  =  3,  21. 

X     y 

6_2^j 

a;     y 

18.  ^-^  =  9,  22. 


a;     y 
7_2 


I-?  =  6. 


.+?= 

y 

3^-?  = 
y 

7 

2' 
.26 
3* 

2x_§  = 

y 

:3, 

8x  +  l*i. 

+  6  = 

1-3. 

=  8, 

^^ 

=  6. 

a; 

=  57, 

^  +  22/ 

=  7. 

SIMULTANEOUS   EQUATIONS.  119 


23.   x-^  +  7  =  Sx-^- 

y                y 

-ll  =  7x  +  21.            24.  ^  +  y  =  2a, 
x-y  =  26. 

26.   ax  +  6y  =  (a  +  6)2, 

32. 

ax  -  6y  =  a2  +  62, 

ax—  by  =  a^  -  b^. 

x  +  y  =  2a. 

26.   ax  +  by  =  a^  +  b% 

33. 

a2x  +  62?/  =  c2, 

bx  +  ay  =  2  ab. 

a^x  +  b^y  =  c3. 

27.  X-  y  =  a-b, 
ax-by  =  2a^-2  b\ 

34. 

ax  +  6?/  =  1, 
6x  +  ay  =  1. 

28.   ax  +  6y  =  a2  -  62, 

6x  +  ay  =  a2  -  62. 

35. 

X     y 

29.    b'h^-ah)  =  (i, 

bx-\-  ay  =  a  +  b. 

X     y 

30.  x  +  y  =  a  +  6,  36.    (a  + 6)x +(a  +  c)y  =  a  +  6, 
ax-by=:b^-  a\  (a  +  c)  x  +  (a  +  6)  y  =  a  +  c. 

31.  bx-ay  =  b^,  87.   (a  + 6)x -(a  -  6)y  =  3a6, 
ax-by  =  a'.  (a  +  6)  y  -  (a  -  6)  x  =  a6. 

108.  Simultaneous  Equations  with  Three  Unknown  Quantities. 
Three  simultaneous  equations  of  the  first  degree,  con- 
taining the  three  unknown  quantities  x,  y,  and  z,  can  be 
solved  in  the  following  manner  : 

Multiply  the  first  and  second  of  the  given  equations 
by  such  quantities  that,  in  the  resulting  equations,  the 
coefficients  of  one  of  the  unknown  quantities,  z  suppose, 
may  be  equal ;  then  by  addition  or  subtraction  we  elimi- 
nate z. 

Then  take  the  first  and  third,  or  the  second  and  third 
of  the  given  equations,  and  eliminate  2;  in  a  similar  man- 
ner. We  thus  obtain  two  simultaneous  equations  con- 
taining the  two  unknown  quantities  x  and  y  j  and  these 
can  be  solved  as  in  Art.  104. 


120  SIMULTANEOUS   EQUATIONS. 

Ex.  1.    Solve  the  equations 

5x  —  42/  +  40  =  9. 
Multiply  the  first  equation  by  2  ;  then 

4x  +  8?/  +  2;2=14. 
Subtract  the  second  equation  ;  then 

x  +  Qy  =  Q (1.) 

Now  multiply  the  first  equation  by  4  ;  then 

8  ic  +  16  ?/ +  4  0  =  28. 
Subtract  the  third  equation  ;  then 

3x  +  20«/  =  19 (ii.) 

From  (i.)  and  (ii.)  we  obtain  ic  =  3,  y  =  I.    Substitute  these 
values  in  the  first  of  the  given  equations  ;  then 

6  +  2  +  0  =  7;  therefore  z=  -\. 
Thus  a;  =  3,  2/  =  1 ,  and  z  =  —\  are  the  values  required. 

Ex.  2.   Solve  the  equations 


^     y    2 

_1_ 

X 

5, 

1+1. 

Z        X 

_1_ 

y 

:3, 

1+1. 

X     y 

z 

:1. 

Add  the  first  two  equations  ;  ■ 

then 

2  1 

-  =  8,  whence  z  =  -' 
z  4 

Add  the  first  and  third  equations  ;  then 

2  1 

-  =  6,  whence  y  =  -• 
y  3 

Add  the  second  and  third  equations  ;  then 

-  =  4,  whence  x  =  — 
a?  2 


SIMULTANEOUS   EQUATIONS.  121 

EXAMPLES   XXV. 

1.  y-\-  z  =  U,  10.          x  +  y-\-z  =  l, 

z-{-x=is,  ^-^y  +  iz  =  h 

x  +  y  =  2^.  2     4 

8..  +  .  =  2a,  f  +  '-f-t'- 

z-\-x  =  2b, 

x  +  y  =  2c.  11.  ax  +  by  =1, 

8.        x  +  y  +  z  =  l,  by  +  cz  =  h 

2x  +  Sy  +  z  =  ^,  cz-^ax=l. 

^x-\-9y-{-z  =  l6.  ^2.  cy-hbz  =  bc, 

4.  5  X  4-  3  y  +  7  0  =  2,  az+  cx  =  ca, 

2  x  -  4  y  +  9  ^  =  7,  bx-\-ay  =  ab. 

Sx  +  2y  +  Gz  =  S. 

4  13.    x  +  y-\--  =  S, 

5x  +  y+3«  =  5,  1 


2x-3;  +  4.  =  20.  2a:  +  3y  + 

x-{-2y-Sz  =  6,  3x-2y  +  -=ll. 
2x  +  4i/-72  =  9,  ^ 

x4-2w  +  32;  =  4,  a6      z 


2x  +  Sy  +  4:Z  =  Q,  2x  +  §i[  +  £  =  2, 

3x  +  4y  +  5iS  =  8.  a         b       z 

8.  3x  +  2y  +  5.  =  21,  3x_2j,^3c^,,^ 
2x-3?/  +  4^  =  ll,  a        0        z 

x-\-Sy  +  7z  =  20.  15.   a;  _  ay  +  a2^  =  qjB^ 

9.  lOx-2y  +  40  =  5,  x-by  +  b^z  =  b^ 

Sx+by-Sz  =  7,  x-cy  +  c''z  =  cK 
x  +  Sy-2z  =  2. 


122  PROBLEMS. 


CHAPTER  IX. 

Problems. 

109.  We  shall  now  give  examples  of  problems  which 
involve  more  than  one  unknown  quantity,  and  in  which 
the  relations  between  the  known  and  unknown  quantities 
are  expressed  algebraically  by  means  of  equations  of  the 
first  degree. 

Many  of  the  problems  given  in  Chapter  VII.  really  con- 
tain two  unknown  quantities,  but  the  given  relations  are 
in  those  cases  of  so  simple  a  nature  that  it  is  easy  to  find 
an  equation  giving  one  of  the  unknown  quantities  in 
terms  of  the  known  quantities,  and  when  one  of  the 
unknown  quantities  is  found,  the  other  is  immediately 
determined. 

Ex.  1.   Find  two  numbers  such  that  the  greater  exceeds  twice 
the  less  by  three,  and  that  twice  the  greater  exceeds  the  less  by  27. 
Let  X  and  y  be  the  numbers,  of  which  x  is  the  greater. 
Then  we  have  by  the  conditions  of  the  problem 
a;  -  2  y  =  3, 
and  2x—    y  =  27. 

Multiply  the  first  equation  by  2  ;  then 
2x-4:y  =  6. 

Now  subtract  the  members  of  this  last  equation  from  the  corre- 
sponding members  of  the  second  equation,  and  we  have 

Sy  =  21,OTy  =  7. 


PROBLEMS.  123 

Then  from  the  first  equation 

x  =  S-\-2y  =  S  +  Uz=n, 
Thus  the  numbers  are  17  and  7. 

Ex.  2.  A  number  of  two  digits  is  equal  to  seven  times  the  sum 
of  its  digits,  and  the  digit  in  the  ten's  place  is  greater  by  four  than 
the  digit  in  the  unit's  place.     What  is  the  number  ? 

Let  X  be  the  digit  in  the  ten's  place,  and  y  be  the  digit  in  the 
unit's  place. 

Then  the  number  is  equal  to  10  a;  +  y,  for  the  x  represents  so 
many  tens  ;  also  the  sum  of  the  digits  is  x  +  y. 

Hence  we  have  10 a;  +  y  =  7(a;  4-  y)j 

that  is  10x  +  2/  =  7aj  +  7y; 

.♦.  3  a;  =  6  y,  or  X  =  2  y. 

We  have  also  x  =  y  +  4. 

Hence  2  y  =  y  +  4,  or  y  =  4, 

and  therefore  x  =  8. 

Thus  the  required  number  is  84. 

Ex.  3.  Find  the  fraction  which  is  equal  to  \  when  its  numerator 
is  increased  by  unity,  and  is  equal  to  ^  when  its  denominator  is 
increased  by  unity. 

Let  X  =  the  numerator  of  the  fraction,  and  y  =  the  denominator. 
Then  we  have  by  the  conditions  of  the  problem 

^+1  =  1,   and     ^     -1 


y        2'  y  +  1     3 

Multiply  the  first  equation  by  y  ;  then 


.+i=|. 


Multiply  the  second  equation  by  y  +  1 ;  then 
a;  =  i(y  +  l). 


124  PEOBLEMS. 

Subtract  the  corresponding  members  of  the  last  two  equations ; 
and  we  have 

from  which  we  jfind  that  y  =  S.     Then,  since  y  is  8, 

2 
Thus  the  fraction  is  f . 

Ex.  4.  A  man  and  a  boy  can  do  in  15  days  a  piece  of  work 
which  would  be  done  in  2  days  by  7  men  and  9  boys.  How  long 
would  it  take  one  man  to  do  it  ? 

Let     X  =  the  number  of  days  in  which  one  man  would  do 
the  whole ; 
and  let  y  =  the   number  of   days  in  which  one  boy  would  do 
the  whole. 

Then  a  man  does  -th  of  the  whole  in  a  day ;  and  a  boy  does 

X 

-  th  of  the  whole  in  a  day. 

y 

Now  by  the  conditions  of  the  problem  a  man  and  a  boy  together 
do  jijth  of  the  whole  in  a  day. 

Hence  we  have  -  +  -  =  — 

X      y      16 

We  have  also,  since  7  men  and  9  boys  do  half  the  work  in  a  day, 

1  +  5  =  1. 

X     y     2 

Multiply  the  first  equation  by  9  and  subtract  the  second ;  then 

2_Z=1_1 
X     X     15     2' 

9         1  1        f1 

thatis  ±=:±-,oT^  =  ^; 

X     10        X     20 

.-.  x  =  20. 

Thus  one  man  would  do  the  work  in  20  days. 


PROBLEMS.  125 

EXAMPLES  XXVL 

1.  A  and  B  have  $50  between  them,  but  if  A  were  to  lose  half 
his  money,  and  B  |  of  his,  they  would  then  have  only  $"20.  How 
much  has  each  ? 

2.  A  number  of  two  digits  has  its  digits  reversed  by  the  addi- 
tion of  9.     Show  that  the  digits  differ  by  unity. 

3.  A  man  bought  8  cows  and  50  sheep  for  $  1125.  He  sold  the 
cows  at  a  profit  of  20  per  cent,  and  the  sheep  at  a  profit  of  10  per 
cent,  and  received  in  all  $  1287.50.  What  was  the  cost  of  each 
cow  and  of  each  sheep  ? 

4.  Twenty-eight  tons  of  goods  are  to  be  carried  in  carts  and 
wagons,  and  it  is  found  that  this  will  require  15  carts  and  12 
wagons,  or  else  24  carts  and  8  wagons.  How  much  can  each  cart 
and  each  wagon  carry. 

6.  A  and  B  can  perform  a  certain  task  in  30  days,  working 
together.  After  12  days,  however,  B  is  called  off,  and  A  finished 
it  by  himself  24  days  after.  How  long  would  each  take  to  do  the 
work  alone  ? 

6.  If  the  numerator  of  a  certain  fraction  be  increased  by  1  and 
its  denominator  diminished  by  1,  its  value  will  be  1.  If  the  nu- 
merator be  increased  by  the  denominator  and  the  denominator  be 
diminished  by  the  numerator,  its  value  will  be  4.  Find  the 
fraction. 

7.  Find  the  fraction  such  that  if  you  quadruple  the  numerator 
and  add  3  to  the  denominator  the  fraction  is  doubled,  but  if  you 
add  2  to  the  numerator  and  quadruple  the  denominator  the  frac- 
tion is  halved. 

8.  The  first  edition  of  a  book  had  600  pages,  and  was  divided 
into  two  parts.  In  the  second  edition  one  quarter  of  the  second 
part  was  omitted  and  30  pages  added  to  the  first.  The  change 
made  the  two  parts  of  the  same  length.  What  were  they  in  the 
first  edition  ? 


126  .  PEOBLEMS. 

9.  If  A  were  to  receive  $  10  from  B,  he  would  then  have  twice 
as  much  as  B  would  have  left ;  but  if  B  were  to  receive  $  10  from 
A,  B  would  have  three  times  as  much  as  A  would  have  left.  How 
much  has  each  ? 

10.  A  farmer  sold  30  bushels  of  wheat  and  50  bushels  of  barley 
for  $93.75.  He  also  sold  at  the  same  prices  50  bushels  of  wheat 
and  30  bushels  of  barley  for  $96.25.  What  was  the  price  of  the 
wheat  per  bushel  ? 

11.  A  rectangle  is  of  the  same  area  as  another  which  is  6  yards 
longer  and  4  yards  narrower ;  it  is  also  of  the  same  area  as  a  third, 
which  is  8  yards  longer  and  5  yards  narrower.     What  is  its  area  ? 

12.  A  and  B  can  together  do  a  piece  of  work  in  15  days.  After 
working  together  for  6  days,  A  went  away,  and  B  finished  it  by 
himself  24  days  after.     In  what  time  would  A  alone  do  the  whole  ? 

13.  An  income  of  $  120  a  year  is  derived  from  a  sum  of  money 
invested  partly  in  3|  per  cent  stofek  and  partly  in  4  per  cent  stock. 
If  the  stock  be  sold  out  when  the  3 1  per  cents  are  at  108  and  the 
4  per  cents  at  120,  the  capital  realized  is  $  3672.  How  much  stock 
of  each  kind  was  there  ? 

14.  A  number  of  two  digits  is  equal  to  seven  times  the  sum  of 
its  digits ;  show  that  one  digit  must  be  twice  the  other. 

16.  Find  all  the  numbers  of  two  digits,  each  of  which  is  equal 
to  four  times  the  sum  of  the  digits. 

16.  $  1000  is  divided  between  A,  B,  C,  and  D.  B  gets  half  as 
much  as  A ;  the  excess  of  C's  share  over  D's  share  is  equal  to 
one-third  of  A's  share,  and  if  B's  share  were  increased  by  $  100, 
he  would  have  as  much  as  both  C  and  D.  Find  how  much 
each  gets. 

17.  A  number  has  two  digits,  of  which  the  second  is  double 
the  first ;  and,  if  the  digits  be  reversed,  the  new  number  exceeds 
the  original  number  by  36  ;  find  the  number. 


PROBLEMS.  127 

18.  A  certain  number  consists  of  two  digits,  and  another  num- 
ber is  formed  from  it  by  reading  it  backwards.  If  the  sum  of  the 
two  numbers  is  99  and  the  difference  is  45,  lind  the  digits. . 

19.  In  a  certain  proper  fraction  the  difference  between  the 
numerator  and  the  denominator  is  12,  and  if  each  be  increased 
by  5,  the  fraction  becomes  equal  to  f .    Find  it. 

20.  The  wages  of  10  men  and  8  boys  for  a  day  amount  to  $  31 ; 
and  four  men  receive  $  5.50  more  than  six  boys.  How  much  does 
each  boy  receive  ? 

21.  A  farmer  has  two  farms,  for  each  of  which  he  pays  a  rent  of 
$  7.60  an  acre,  and  his  total  rent  is  $  3375.  If  the  rent  of  one 
farm  were  reduced  by  $  1.25  an  acre,  and  that  of  the  other  by 
$  2.50  an  acre,  his  rent  would  be  $  2500.  What  is  the  acreage  of 
each  of  the  farms  ? 

22.  A  man  has  one  pound's  worth  of  silver  in  half-crowns, 
shillings,  and  sixpences  ;  and  he  has  in  all  20  coins.  If  he  changed 
the  sixpences  for  pennies,  and  the  shillings  for  sixpences,  he  would 
have  73  coins.    How  many  coins  of  each  kind  has  he  ? 

23.  The  price  of  a  passenger's  ticket  on  a  French  railway  is 
proportional  to  the  distance  he  travels  ;  he  is  allowed  25  kilo- 
grammes of  luggage  free,  but  on  every  kilogramme  beyond  this 
amount  he  is  charged  a  sum  proportional  to  the  distance  he  goes. 
If  a  journey  of  200  miles  with  50  kilos,  of  luggage  cost  25  francs, 
and  a  journey  of  150  miles  with  35  kilos,  cost  1C|  francs,  what 
will  a  journey  of  100  miles  with  100  kilos,  cost  ? 

24.  A  has  twice  as  many  dimes  as  dollars ;  B,  who  has  60  cents 
more  than  A,  has  twice  as  many  dollars  as  dimes ;  together  they 
have  one  more  dime  than  they  have  dollars.   How  much  has  each  ? 

25.  Of  the  candidates  in  a  certain  examination,  one-quarter  fail. 
The  number  of  marks  required  for  passing  is  less  by  2  than  the 
average  marks  obtained  by  all  the  candidates,  less  by  11  than  the 
average  marks  of  those  who  pass,  and  equal  to  double  the  average 
marks  of  those  who  fail.  How  many  marks  are  required  for 
passing  ? 


128  MISCELLANEOUS   EXAMPLES   II. 

MISCELLANEOUS  EXAMPLES    IL 

A.  1.   Find  the  value  of 

ab  +  2hc-Zcd      gg  +  &^  -  c3 
6+c  +  d  a2  +  62_c2' 

when  a.  =  1 ,  &  =  2,  c  =  —  3,  and  cZ  =  0. 


2.  Subtract  2a-3  («-&-«)  from  2  6  -  3  (&  -  «  -  6). 

3.  Show  that 

(a^+3)(y+3)-3(a;+l)(2/+l)+3(a^-l)(y-l)-(a:-3)(2/-3)=:0, 
and  that 

(x+2)(^/+2)-4(.'c+l)(y+l)+6a;2/-4(x-l)(2/-l) 

+  (x-2)(i/-2)=0. 

4.  Divide  x^  -  5x*  +  7 x^  -  x^  -  4a;  +  2  by  x^  -  3a:2  +  3a;  _  i. 

5.  Solve  the  equations : 

,.  .    a;a;  +  la;  +  2a;  +  3^^ 
^■^2345 

(ii.)   f  +  |  =  5,  2a;+|-17  =  0. 
3      5  3 

(iii.)   aaj  —  &y  =  a'^  —  62 ^ 
6x-a?/=62_a2i    . 

6.  A  number  of  marbles  were  divided  among  thi*ee  boys^so  that 
the  first  boy  had  10  less  than  half  of  the  whole,  the  second  had 
10  more  than  a  quarter  of  what  was  left,  and  the  third  had  20. 
How  many  marbles  were  there  ? 

B.  1.  Find  the  numerical  value  of 

(a  -  6)2 +  (6  -  c)2  +(c  -  a)2  +(6  -c){c-  a)(a  -  6), 

when  a  =  1,  6  =  2,  and  c  =  ^. 

2.  From  the  sum  of  (2  a  -  6)2  and  (a  -  2  6)2  take  the  square 
of  2(a-6). 


MISCELLANEOUS   EXAMPLES   II.  129 

3.  Show  that 

n^  =  n(n-  l)(n  -  2)+ 3w(n  -  1)+ n, 
and     n4=n(n-l)(w-2)(w-3)+6w(n-l)(w-2)  +  7  7i(n^l)  +  n. 

4.  Find  the  algebraical  expression  which,  when  divided  by 
x2  —  2  a;  +  1 ,  gives  a  quotient  x^  +  2  x  +  1  and  a  remainder  x  +  1. 

5.  Solve 

(i.)  3(x  +  3)2  +  5(x  +  5)2  =  8(x  +  8)2. 

(ii.)   ^(x  +  y)-K^-y)=S  \ 

cm.)  1  +  4^1,    Z_8^1. 
X     y  X     y     6 

6.  A  man  paid  a  bill  of  £100  with  sovereigns  and  crowns, 
using  in  all  130  coins.     How  many  coins  of  each  sort  did  he  use  ? 


C.  1.    Simplify  a  -[a  -^  b  -  {a  -{-  b  ■}-  c  -  a  -\-  b  +  c  +  d}']. 

2.  Find  the  sum,  the  difference,  and  the  product  of  a  +  6  and 
a  _b 

2     2 

3.  Show  that 

ab  -  3(a  -  1)(6  -  1)  +  3(a  -  2)(6  -  2) -(a  -  3) (6  -  3)  =  0, 
and  that 
ab  -  4(a  -  1)(6  -  1)+  6(a  -  2)(6  -  2) 

-  4(a  -3)(b-  3)4-(a  -  4)(6  -  4)=  0. 

4.  Divide  a^ + 6^  +  c^  -  3  a6c  by  a + 6  +  c,  and  from  the  result  write 
down,  without  division,  the  quotient  when  Sx^  -\-Sy^  +  ^  —  12  xyz 
is  divided  hy  2x-\-  2y  -\-  z. 

5.  Solve  the  equations : 

(i.)  x  =  K5?/  +  2),  2/  =  K^-l)- 
(ii.)  ax  -\-  by  =  a^,  bx  +  ay  =  b^. 

6.  A  and  B  each  shoot  30  arrows  at  a  target.  B  makes  twice 
as  many  hits  as  A,  and  A  makes  three  times  as  many  misses  as  B. 
Find  the  number  of  hits  and  misses  of  each. 

X 


130  MISCELLANEOUS   EXAMPLES   II. 

D.  1.  Find  the  value  of 

abc(ab  +  bc  +  cd  -i-  da)  -=-  (a  +  6)  (a  +  c) (a  +  d), 
when  a  =  1,  &  =  3,  c  =  —  5,  and  d  =  0. 

2.  Simplify  (a  +  6)2  -  [2  a^  -  {(a  -  6) (a  +  2  &)  -  6(a  -  6)}]. 

3.  Find  the  continued  product  of  x  +  a,  x  +  b,  and  x  +  c ;  and 
from  the  result  write  down  the  continued  product  of  a  —  x,  a  —  y, 
and  a  —  z. 

4.  Divide  76a%^c^  -  Iba^b^c^  by  ba'^bc^,  and 

a2  _  2  62  _  6  c2  4-  a6  -  ac  +  7  6c  by  a  -  6  +  2  c. 

5.  Solve  the  equations : 

(i.)  ^(ix-l)-iix-2)+l(x-S)  =  0. 

Cii  ^   ^~^  =  y -^  =  a;  +  y 
^  '^       2  3  9 

(iii.)   §  +  ^=22,  l-i  =  20. 
X     y  X     y 

6.  A  square  grass  plot  would  contain  69  square  feet  less  if  each 
side  were  one  foot  shorter.    How  many  square  feet  does  it  contain  ? 

E.  1.    Simplify 


b^  +  {b(a  —  c)+  ac]  —  a{b  —  a  -  c]  +  b(c  -  b  r-  a)  —  ab. 

2.  From  what  must  the  sum  of  3  a^  +  2  a6  —  ac,  3  6^  -j.  2  6o  —  a6, 
and  3  c2  +  2  ca  —  6c  be  taken  in  order  that  the  remainder  may  be 

a2  +  62  +  c2  ? 

3.  Show  that  (x  +  y)2  +  (y  +  zy  +  (0  +  xy 

+  2(a;  +  y)<ix+  z)+  2(y  +  zXv  +  x)+2{z  +  x){z  +  y) 
=  4(x  +  y  +  zy. 

4.  Divide  ^x^  +  -\^  x"^  -  ^x  +  ^  hj  ^x  +  2,  and 

acx^  +  {ad  -  bc)x^  —  (ac  +  bd)x  +  be  by  ax  —  6. 


MISCKLLANEOUS   EXAMPLES   n.  131 

5.  Solve  the  equations : 

(i.)    (a;-l)(x  +  2)  =  a;2  +  3. 


+  .2?/  =  1.8J 


(ii.)    2x-\-  Ay  z=  1.2 
bx 


(iii.)     (a-6)x  =  (a+ 6)y) 
x-\-y  =  2a  / 

6.  Find  two  numbers  such  that  twice  their  difference  is  greater 
by  unity  than  the  smaller  number,  and  is  less  by  two  than  the 
larger  number. 

F.  1.  Find  the  value  of  ^(x -\- y  ■\- zy  -  (t? -\- y^  +  z^) ,  when 
jc  =  3,  ?/  =  —  5,  and  z  =  1. 

2.  Find  the  continued  product  of  a*  +  x*,  a^  +  x^,  a  +  x,  and 
a  —  x. 

3.  Show  that  (1  -  a)(l  -  6)  +  a(l  -  6)  +  &  =  1, 

and  that  (1  -  «)  (l  _  6)  (1  _  c)  +  a(l  -  b)  (1  -  c)  +  6(1  -  c)  +  c  =  1. 

4.  Divide        4y*  -  9y^  +  6y  -  I  hy  2y^ -^  Zy  -  1. 
Divide  also  a^  +  2  62  -  3  c2  +  &c  +  2  ac  +  3  a6  by  a  +  6  -  c. 

5.  Solve  the  equations  : 

(i.)    (a  +  x)(6  +  x)  =  (c  +  x)(d  +  x). 
(ii.)   Sx~7y  =  4,  7x-9y  =  2. 

6.  If  12  lbs.  of  tea  and  3  lbs.  of  coffee  cost  ^8.76,  and  12  lbs. 
of  coffee  and  3  lbs.  of  tea  cost  $6.24,  what  is  the  price  per  lb. 
of  tea  and  of  coffee  ? 


132  FACTORS. 


CHAPTER  X. 

Factoes. 

110.  Definitions.  An  algebraical  expression  which  does 
not  contain  any  letter  in  the  denominator  of  any  term  is 
called  an  integral  expression. 

Thus  a  +  h  and  \  a^  _  i  ^2  are  integral  expressions. 

An  expression  is  said  to  be  integral  with  respect  to  any 
particular  letter  when  that  letter  does  not  occur  in  the 
denominator  of  any  term. 

Thus  —  4 is  integral  with  respect  to  x. 

a      a  +  h 

An  expression  is  said  to  be  rational  when  none  of  its 
terms  contain  square  or  other  roots. 

111.  In  the  present  chapter  we  shall  show  how  factors 
of  rational  and  integral  algebraical  expressions  can  be 
found  in  certain  simple  cases. 

In  arithmetic  we  mean  by  the  factors  of  a  number  its 
integral  divisors  only;  and  similarly,  by  t\\Q  factors  of  an 
algebraical  expression,  we  mean  the  rational  and  integral 
expressions  which  exactly  divide  it. 

112.  Monomial  Factors.  When  any  letter,  or  group  of 
letters,  is  common  to  all  the  terms  of  an  expression,  each 
term,  and  therefore  the  whole  expression,  is  divisible  by 
that  letter,  or  group  of  letters. 


FACTORS.  133 

Thus  ab  +  ac  =  a(^b  +  c), 

a'^b  +  a62  =  ab(a  +  6), 
and  3  ay^y  +  6  axY  =  3  0x2^/  (x  +  2  y2) . 

Monomial  factors,  if  there  are  any,  can  be  seen  by- 
inspection,  and  the  whole  expression  can  be  at  once  writ- 
ten as  the  product  of  the  monomial  factor  and  its  co-factor, 
as  in  the  above  examples.  In  what  follows  therefore  we 
need  only  consider  expressions  which  have  been  freed 
from  monomial  factors. 

EXAMPLES  XXVII. 
Find  the  factors  of  the  f ollowmg  expressions : 

1.  x^  +  X.  7.   ic*  -bx^y-\-  20x2^/2. 

2.  a2  -  ab.  8.   a^  -  a*x  +  a2ic2. 

3.  ab  -  be.  9.   6  ax*  -  5  a^x^  +  20  a2x3. 

4.  2  ax  -I-  ^  x2.  10.   3  a^x*y^  -  ^  oh^yK 

5.  4x8-3x2.  11.   Ila25c2- ia&2c3. 

6.  a8-3a26.  12.  p'^q^r^  -  7  j^q^. 

113.  factors  found  by  comparing  with  Known  Identities. 
Sometimes  an  algebraical  expression  is  of  the  same  forin 
as  some  known  result  of  multiplication :  in  this  case  its 
factors  can  be  written  down.  We  proceed  to  apply  this 
principle  in  the  case  of  the  most  important  forms  of 
algebraical  expressions. 

114.  We  know  that 

and  a^  -  2a6  +  62  =  (a  -  bf. 


134        -  FACTORS. 

Hence,  when  a  trinomial  expression  consists  of  the  sum 
of  the  squares  of  any  two  quantities  plus  tlvice  the  product 
of  the  quantities,  it  is  equal  to  the  square  of  their  sum. 
And  when  a  trinomial  expression  consists  of  the  sum  of 
the  squares  of  any  two  quantities  minus  tivice  their  product, 
it  is  equal  to  the  square  of  their  difference. 

Hence  it  is  easy  to  recognize  when  a  trinomial  is  a 
complete  square ;  for  two  of  the  terms  must  be  squares, 
and  the  remaf~ang  term  must  be  twice  the  product  of  the 
quantities  v^  ares  are  the  other  terms. 

The  two  tt  3h  are  squares  must  both  have  the 

same  sign. 

Thus  a2  +  6  a&  +  9  b'^,  that  is  a"^  +  (3  &)2  +  2  a  (3  &),  consists  of 
the  squares  of  a  and  3  b  together  with  twice  the  product  of  a  and 
3  b,  and  hence 

a2  +  6a6  +  962=  (a  +  3&)2. 
Again, 
^ab  -  a^  -  4:b'^  =  - (a"^  +  4b^  -  i ab)  =  -  {a"^  +  (2  6)2  -  2 a  (2 6)} 

=:-(a-2&)2. 
As  other  examples,  we  have 

lea*  4-  8a-^  +  1  =(4a2)2  +  2  (4^2)  +  i  =  (4a2  +  1)2, 
ics  ^  4  x^y^  +  4  icy*  =  a;  (x2  -  4  xy^  +  4  y*)} 

=  x{x^  -  2  a;  (2^/2)+  (2?/2)2}  =  x  (x  -2y'^)\ 
and  (a  +  6)2  -  2c  (a  +  &)+  c2  =  {(a  +  b)-cf. 

Note.  —  We  may  consider  that  a^  —  2ab-\-b'^  is  equal  to 
(b  -  a)2  instead  of  (a  -  6)2,  for  by  the  Law  of  Signs  (&  -  a)\ 
that  is  {-{a  -  b)f,  is  equal  to  (a  -  6)2.  In  fact,  when  we  find 
that  an  expression  is  equal  to  the  product  of  two  factors,  we  may 
equally  well  consider  it  is  as  the  product  of  the  same  two  factors 
with  all  the  signs  changed. 


I 


FACTORS.  135 

EXAMPLES  XXVIII. 
Find  the  factors  of  the  following  expressions : 

1.  4x2  +  4a;+l.  9.   3a2  +  6a6  +  362. 

2.  9x^-Qx+  1.  10.   5a4  _  lOa^b  +  5  b^ 

3.  l-8x2  +  16x*.  11.   a»-6a%-{-9ab^. 

4.  4a2_i2a6  +  962.  12.   3  a^  -  30  a*68  +  75  a^fe^ 

5.  9 a*  +  24 a262  +  16 6*.  13.  4^22^2   .a4_4y4. 

6.  x2  +  xy  +  iy2.  14.   8x2  ^,.^^^-4. 

7.  4  a2x2  +  4  a6xy  +  &2y2.  15.   4  ^i/S Ja'J4  x22/2  +  x^y. 

8.  25a*x^-S0a'^b^xy  +  9b*y^.  16.   x22/2  +  ^^y  +  ^  xy8. 

17.  (a  +  6)2  +  4c(a  +  6)+4c2. 

18.  (X2  +  2/2)2  _  2  (X2  +  y')  02  +  ;j4. 

19.  4x2y2  +  4  (a  +  b)  xy  +  (a  +  b)\ 

20.  9  (a  +  6)2  _  6  c  (a  +  6)  +  c2. 

115.   From  the  formula 

a'-b'  =  (a  +  b){a-b), 

we  see  that  the  difference  of  the  squares  of  any  two  quan- 
tities is  equal  to  the  product  of  the  sum  and  the  difference 
of  the  quantities.  ^ 

Thus  a2  -  4  &2,  or  a^  -  (2  6)2,  is  equal  to  (a  +  2  6)  (a  -  2  &). 
Also  9x8  -  4xy2,  that  is  x{(3x)2  -  (2  yY);  is  equal  to 

x(3x  +  2i/)(3x-22/). 
As  other  examples,  we  have : 

8  axy  -  18  a^x^y^  =  2  axy  (4  -  9  a^xY')  =  2  axy  {22  -  (3  axyy] 
=  2  axy  (2  -  3 axy)  (2  +  3 axy), 
a*  -  6*  =  (a2  +  &2) (^2  _  52)  ^  (aj2  +  ^,2) (a  +  5)  («  _  5)^ 
and         792  -  712  =  (■yg  +71) (-79  _  71) ^  150  x  8  =  1200. 


136  FACTORS. 

We  may  deal  with  the  squares  of  multinomial  expres- 
sions in  precisely  the  same  way  as  with  the  squares  of 
monomial  expressions. 

Thus  (a  +  by  -  c^  =  {{a  +  6)  +  c]  {{a  +  6)  -  c} 

=  (a  +  6  +  c)(a  +  &-c), 

ai  +  a252  +  54  =  (^2  +  52)2  _  («5)2  ^  («2  +  52  _^  «5)  (^2  +  52  _  <25)^ 

and    (2a- 6  +  2c)2-(a  +  46 +  c)2 

=  {(2  a  -  &  +  2  c)  +  (a  -f-  4  6  +  c) }  {(2  a  -  6  +  2  c)  -  (a  +  4  6  4-  c)} 

=  (3 a  +  3 6  +  3 c)(a  -  5 6  +  c). 

EXAMPLES  XXIX. 
Find  the  factors  of  the  following  expressions : 


1. 

a2  -  9. 

19. 

X*  -  2/4. 

2. 

16  -  62. 

20. 

X*  -  16  yK 

3. 

25  a2  _  62. 

21. 

81  a*  -  16  6*. 

4. 

a;2  -  9  y2. 

22. 

625  a* -256x4. 

5. 

16x2 -9  2/2. 

23. 

x*^/*  -  a*64. 

6. 

64  a2  _  49  62. 

24. 

a464  _  81  c4#. 

7. 

4a2-81&2. 

25. 

81  xV  -  1. 

8. 

x2-9y*. 

26. 

16  a464c4  _  1. 

9. 

36  a* -49  62. 

27. 

16-81xV, 

10. 

4a262_9c2. 

28. 

X8  -  ^8. 

11. 

9«2x2_49622/2. 

29. 

a8  _  68^8. 

12. 

49a262c2_36x2i/2«2. 

30. 

aio  -  a2. 

13. 

4x2/2-9x3. 

31. 

{a  +  6)2  -  c2. 

14. 

8  a62  _  18  aK 

32. 

(a  +  6)2-4c2. 

15. 

3  «&  -  108  aK 

33. 

4  (X  +  2/)2  -  1. 

16. 

7  a^  -  28  a9. 

34. 

9  (X  -  y)2  -  4. 

17. 

8  x^y  -  32  xy^. 

35. 

(x  +  2/)2-(x-2/)2. 

18. 

7a6c2-7a868. 

36. 

(2a  +  6)2-(26  +  a)2. 

FACTORS.  137 

37.  x^-(x-yy.  41.  (a2  +  52)2  _  4  <2262. 

38.  a2  _  (2  6  _  a)2.  42.  (a  +  5  +  c)2  -  (a  -  6  -  c)2. 

39.  4(a  +  &)2-(a-6)2.  43.  (3a +6_  2c)2-(<i+3  6-c)2. 

40.  9  (a;  +  2/)2  -  4  (X  -  y)2.  44.  (a  -  2  6  +  3c)2 -(a  -  c)2. 

46.    (3  x2  +  a;  -  2)2  _(a;2  _  a;  _  2)2. 

116.   From  the  formulae 

/■ 

a^  +  a^  =  (a;  +  a)  (a^  —  aa;  +  a^), 
and  x^-a^  =  {x-a){a?-\-ax  +  a-),      [Art.  87.] 

we  see  that  the  sum  of  the  cubes  of  any  two  quantities  is 
divisible  by  the  sum  of  the  quantities ;  and  that  the  dif- 
ference of  the  cubes  of  any  two  quantities  is  divisible  by 
the  difference  of  the  quantities. 

Thus  8  a3  4.  27  68,  that  is  (2  «)»  +  (3  by,  is  divisible  by  2  a  +  3  6, 
and  the  quotient  is  (2  aY  -  (2  a)  (3  h)  +  (3  6)2,  that  is 
4  a2  _  c  a&  +  9  h\ 
Also  27  x^  -  8,  that  is  (3a;)»  -  2^,  is  equal  to 

(3ic-2){(3:?)2  +  (2.3x)+22}  =  (3x-2)(9x2  +  6x4-4). 
As  other  examples, 

asjs  _  1  c8  =(a6  _  ^c)(a262  +  ^  a&c  +  ic2), 
and  a6  _  56  _  (fljs  +  68)  (^^a  _  ^,8) 

.  =(a  +  6)(a2  -ab  +  b^)(a  -  b)(ia^  +  ab  +  62). 

The  cubes  of  multinomial  expressions  may  be  dealt 
with  in  precisely  the  same  way. 

Thus  (a  +  6)3  -  c3  =  {(a  +  6)  -  c}{(a  +  6)2  +  (a  +  6)  c  +  c2}, 
B,nd  (x-2yy-(y-2xy 

=  (x  -  2y  -  y  -  2x){{x  -  2yy  -\-(x  -2y)(y  -  2x)-^(y  -  2xr-} 
=  (3a;-3y)(3a;2-3a:y  +  3!/2). 


188  FACTORS. 

EXAMPLES  XXX. 
Find  factors  of  each  of  the  following  expressions : 

1.  a3_8  63.  11.  S  a^b  + 24:  ab\ 

2.  8a^  +  63.  12.  40  a^bc  -  5  fe^c*. 

3.  8a3_i25ic8.  13.  a^  -  64. 

4.  a3- 125x6.  14.  64  a'- -729  66. 

5.  4a3  +  32  63.  15.  x^^  -  a^¥. 

6.  27x3-i?/3.  16.  (x-i-2yy-y^ 

7.  x3y3  +  . ^1^0^353.  17.  (x  +  2yy  +  Cy  +  2xy, 

8.  8  a666  +  x^.  18.  (2  y  -  x)^  -  (2  x  -  i/)3. 

9.  2xy^-l  xK                         19.  (x  -  3  ?/)3  -  (y  -  3  x)^. 
10.   9a*62-ia6^                      20.  (2?/ -  x)^ +  (2x  -  y)3. 

117.   By  multiplication  we  have 

(a?  +  a)  (»  +  6)  =  a^  +  (a  +  6)  ic  +  ab. 

Hence  conversely,  if  an  expression  of  the  form 
a^ -\- px -\- q  he  the  product  of  the  two  factors  x  +  a  and 
x-\-b,  the  given  expression  must  be  the  same  as 

a^  +  (a  4-  6)  a;  +  a6 ; 

we  must  therefore  have  p  =  a  +  b  and  q  =  ab.     Hence 
a  and  b  are  such  that  their  sum  is  p,  and  their  product  q. 

For  example,  to  find  the  factors  of  x^  +  7  x  +  12.  The  factors 
will  be  X  +  a  and  x  +  6,  where  ab  =  12  and  a  4-  6  =  7.  Hence  we 
must  find  two  numbers  whose  product  is  12  and  whose  sum  is  7. 
Pairs  of  numbers  whose  product  is  12  are  12  and  1,  6  and  2,  and 
4  and  3  ;  and  the  sum  of  the  last  pair  is  7.  Hence 
x2  +  7  X  +  12  =  (x  +  4)  (x  +  3). 

Again,  to  find  the  factors  of  x^  —  7  x  +  10,  we  have  to  find  two 
numbers  whose  product  is  10,  and  whose  sum  is  —  7.  Since  the 
product  is  +  10,  the  two  numbers  must  both  be  positive  or  both 
negative ;  and  since  the  sum  is  —  7,  they  must  both  be  negative. 


FACTORS.  139 

The  pairs  of  negative  numbers  whose  product  is  10  are  —  10  and 
—  1,  and  —  5  and  —  2  ;  and  the  sum  of  the  last  pair  is  —  7.     Hence 

a;2_7x+10=(x-5)(a;-2). 

Again,  to  find  the  factors  of  x^  +  3  x  —  18,  we  have  to  find  two 
numbers  whose  product  is  —  18,  and  whose  sum  is  3.  Since  the 
product  is  —  18,  one  of  the  numbers  is  positive  and  the  other 
negative.  The  pairs  of  numbers  whose  product  is  —  18  are  —  18 
and  1,-9  and  2,-6  and  3,-3  and  6,-2  and  9,  and  —  1  and 
18  ;  and  of  these  pairs  the  sum  of  6  and  —  3  is  3.     Hence 

x2  +  3  X  -  18  =  (x  +  6)  (x  -  3). 

EXAMPLES  XXXI. 
Find  the  factors  of  each  of  the  following  expressions : 

1.  x2  +  4x  +  3.  13.   x2  +  6x-14. 

2.  x2-4x  +  3.  14.  X2-X-132. 
8.  x2-6x  +  8.  16.   x2  +  18x  +  72. 
4.   x2-8x  +  15.  16.   x2-5x-84. 

6.  x2  -  11 X  +  18.  17.  x2  -  25  X  +  150. 

6.  x2 +  9x4-20.  18.  x2  +  5x-150. 

7.  x2  +  2x  -  3.  19.  x2  +  11 X  -  180. 

8.  x^  +  ix-5.  20.  X2-X-156. 

9.  a:2  +  x-6.  21.  x2-31x  +  240. 

10.  x2  -  X  -  6.  22.  x2  -  17  X  -  200. 

11.  x2  4-2x-35.  23.   x2-34x  +  288. 

12.  x2-3x-10.  24.   x2-35x-200. 

118.    By  multiplication  we  have 

{ax  +  b)  {ex  -\-d)  =  ac3?  -}-  {ad  -{■hc)x-\-  bd. 

Hence  conversely,  if  an  expression  of  the  form 
px^  -{-qx-{-r  be  the  product  of  the  two  factors  ax-\-  b 
and  ex  -f  d,  the  given  expression  must  be  the  same  as 


140  FACTOES. 

acx^  -f  {ad-\-  bc)x  +  bd-,  we  must  therefore  have  ac  =p, 
hd  =  r,  and  ad-{-hc  =  q.  We  can  in  simple  cases  find  by 
trial  the  values  of  a,  b,  c,  d  which  satisfy  these  relations. 

For  example,  to  find  the  factors  of  3x^  —  16x+  6.  The  Sx^ 
can  only  be  given  by  the  multiplication  otSx  and  .r.  The  5  could 
only  be  given  by  the  multiplication  of  5  and  1,  or  —  5  and  —  1 ; 
and,  since  the  middle  term  is  negative,  the  latter  pair  must  be 
taken.  We  have  now  only  a  choice  between  (3  oj  —  5)  (ic  —  1)  and 
(3x  —  l)(a:  —  5),  and  it  is  at  once  seen  that  the  latter  must  be 
taken  in  order  that  the  coefficient  of  x  in  the  product  may  be  —  16. 
Thus  the  factors  required  are  3  ic  —  1  and  cc  —  5. 

Again,  to  find  the  factors  of  5  x^  +  32  a;  —  21,  the  6x^  can  only 
be  given  by  the  multiplication  of  5  x  and  x.  The  end  term  —  21 
can  be  the  product  of  —21  and  1,-7  and  3,  —3  and  7,  or  —1  and 
21.  The  possible  pairs  of  factors,  so  far  as  the  two  end  terms 
are  concerned,  are  therefore  (5x  =F  21)(x  ±  1),  (5x  T  7)(x  i  3), 
(5 X  T  3)  (x  ±  7),  and  (5  x  T  1)  (ic  ±  21).  It  will  be  found  that  of 
these  pairs  the  one  which  will  give  32  x  for  the  middle  term  is 

(5x-3)(x+7). 


119.  The  factors  of  oc^  —  5  xy -\- 4:y^  can  be  found  in 
the  same  way  as  the  factors  of  cc^  —  5  ic  +  4.  For  we 
must  find  two  quantities  whose  product  is  4  y^,  and  whose 
sum  is  —  5y:  these  are  —  4 2/  and  —  y.     Hence 

a^  —  5  xy  +  4:y^  =  (x  —  4:y)  (x  —  y). 

So  also  the  factors  of  3x^  —  16  xy  -\-  5y^  can  be  found  in 
the  same  way  as  the  factors  of  3  o^  —  16  a;  +  5 ;  and  the 
factors  of  either  can  be  written  down  as  soon  as  the  fac- 
tors of  the  other  are  known. 

For  example,  if  we  know  that  5  x^  +  32  x  -  21  =  (5  x  -  3)  (x  +  7), 
we  must  have  6  x^  4-  32  xj/  —  21 2/2  =  (5  x  -  3  y)  (x  +  7  y). 


FACTORS.  141 

EXAMPLES   XXXII. 
Find  the  factors  of  each  of  the  following  expressions : 

1.  3a;2-10a;  +  3.  20.    7x2  +  123x-54. 

2.  3a;2-17x+10.  21.   24x2-30x-75. 

3.  2x2+ 11  x  + 12.  22.   a:2  +  4x?/  +  3t/2. 

4.  2x2  +  3x  -  2.  23.   x*  _  6xy  +  8?/2. 

5.  3x2  + 7x- 6.  24.  x2  -  llxy  +  I82/2. 

6.  4x2  + x-3.  25.   x2  +  5xi/-14y2. 

7.  6x2-38x  +  21.  26.  x2  -  25 xy  +  150 y2. 

8.  3x2  +  llx-20.  27.    x2-35xy-2002/2. 

9.  7  x2  -  33x  -  54.  28.  3x2  -  17  xy  +  10 y\ 

10.  6x2-38x  +  48.  29.  7x2  -  33xy  -  541/2 

11.  7x2  +  75x-108.  30.  24  x2  -  70  xy  -  75  y2. 

12.  9x2  +  130x-75.  81.  x*- 13x2 +  36. 

13.  4  x2  +  21 X  -  18.  32.  X*  -  25  x2y2  +  144  y*. 

14.  4x2  +  4x-15.  33.  36 x*  -  97 x2y2  +  36 y*. 

15.  6x2+55x-50.  34.  x3-3x2-18x. 

16.  10  x2  +  3  X  -  1.  86.  x^y  -  x2y2  _  2  xy*. 

17.  132x2  + x-1.  86.  15x4?/-4xV-4x2y8. 

18.  4x2-5x+l.  37.  75x2/3 -130x2?/4_  9  a;8y5. 

19.  12x2  +  50x-50. 

120.  It  is  clear  that  the  method  of  finding  by  trial  the 
factors  of  an  expression  of  the  form  pa?  +  ga;  +  r,  where 
p,  q,  r  are  known  numbers,  would  be  very  tedious  if 
there  were  many  pairs  of  numbers  whose  product  was  p, 
and  msiiij  pairs  whose  product  was  r,  for  there  would 
then  be  very  many  pairs  of  factors  which  would  agree 
with  the  given  expression  so  far  as  the  end  terms 
were  concerned,  and  out  of  these  the  single  pair  which 
would  give  the  correct  middle  term  would  have  to  be 


142  FACTORS. 

sought.  It  would,  for  example,  be  almost  impossible  to 
find  the  factors  of  2310  x"  -  2419  x  —  9009  in  this  way. 

Again,  we  not  unfrequently  meet  with  such  an  expres- 
sion as  a;^  +  6  a;  +  7  which  cannot  be  written  as  the  prod- 
uct of  two  factors  altogether  rational,  and  in  such  a  case 
it  would  be  impracticable  to  try  to  guess  the  factors. 

We  therefore  need  some  method  of.  finding  the  fac- 
tors of  a  quadratic  expression  which  is  applicable  to  all 
cases.     This  method  we  proceed  to  investigate. 

121.  We  first  note  that  since  a^  -f  2  aa;  -f  a^  is  a  perfect 
square,  namely  {x-\-aY,  it  follows  that  in  order  to  co^^ 
plete  the  square  of  which  x^  and  2aa;  are  the  first  two  terms, 
we  must  add  the  square  of  a ;  that  is,  we  must  add  the 
square  of  half  of  the  coefficient  of  x. 

For  example,  x^  +  6  a;  is  made  a  perfect  square  by  the  addition 
of  (1)=^ ;  and  the  square  \sofi  +  Qx  +  9,  which  is  (x  +  3)2. 

So  also,  Qci'^  -\-  hx  is  made  a  perfect  square  by  the  addition  of 
(1)2 ;  and  the  square  is  x^  +  5  aj  +  (-\^),  which  is  (x  +  f  )2. 

And  x2  —  7  X  is  made  a  perfect  square  by  the  addition  of  (-  |)2, 
that  is  of  \^  ;  and  the  square  is  x^  —  7  x  +  Y,  which  is  (x  —  |)2. 

Whatever  a  may  be,  x^  -|-  ax  is  made  a  perfect  square  by  the 

addition  of  (  -  i  ,  f  or 


•      / 
1212.   We  will  now  find  the  factors  of  ax^  -{-hx-\-c  by 

a  method  which  is  applicable  in  all  cases. 

The  problem  before  us  is  to  find  two  factors  which  are 
to  be  rational  and  integral  with  respect  to  x,  and  are  there- 
fore of  the  first  degree  in  a;,  but  which  are  not  necessarily 


FACTORS.  143 

rational  and  integral  with  respect  to  arithmetical  num- 
bers or  any  other  letters  which  may  be  involved. 
We  first  apply  the  method  to  some  examples. « 

Ex.  1.    Find  the  factors  of 

a;2  +  6  X  +  8. 

As  we  have  seen  in  the  last  article,  x"^  +  6x  is  made  a  perfect 
square  by  the  addition  of  3'^ ;  hence,  adding  and  subtracting  3^,  we 
have 

x^-\-6x  +  S  =  x^  +  ex  +  9-9  +  S 

=  (x  + 3)2-1. 

The  last  form  of  the  expression  is  the  difference  of  two  squares, 
and  the  factors  are  therefore  x  +  3  +  1  and  x  +  3  —  1,  that  is  x  +  4 
and  X  +  2.    Hence  the  factors  of  x^  +  6  x  +  8  are  x  +  4  and  x  +  2. 

Ex.  2.   Find  the  factors  of  x^  -  3  x  -  28. 

x2  —  3  X  is  made  a  perfect  square  by  the  addition  of  (—  |)2,  that 
is  of  I ;  hence,  adding  and  subtracting  |,  we  have 

a;2  -  3x  -  28  =  x2  -  3x  +  I  -  f  -  28 

=  (X-3)2_121 

=  (a;_|)2_(Y)2. 
Now  the  factors  of   the  last  expression  are  x  —  f  +  -^  and 

a^  -  f  -  ¥• 

Thus  the  required  factors  are  x  +  4  and  x  —  7. 

Ex.  3.  To  find  the  factors  of  x2  +  6  x  +  7. 
x2  4-  6  X  is  made  a  perfect  square  by  the  addition  of  3^ ;  hence 
we  write 

x2  +  6x+7  =  x2  4-6x  +  9-9  +  7 

=  (x  +  3)2  -  2 

=  (x  +  3)2-(v2)2. 

Now  the    factors    of    the  last   expression,   and   therefore    of 

x2  +  6  X  +  7,  are  X  +  3  +  V2  and  x  +  3  -  ^2. 


144  FACTORS. 

Ex.  4.    To  find  the  factors  of  3  ^2  -  10  a;  +  3. 

3x2  -  lOx  +  3  =  3  (x2  -  -V*x  +  1). 

Now  x2— ^-x  is  made  a  perfect  square  by  the  addition  of 
(-  f)2,  that  is  of  -%^- ;  hence 

x2  _  -ijO  X  +  1  =  x2  -  -i/x  4. 2^  _  2^5  4.  1 

The  factors  of  the  last  expression  are  x  —  f  +  f  and  sc  —  |  —  f , 
that  is  X  —  1^  and  x  —  3  ;  hence 

3x2  -  lOx  +  3  =  3  (x  -  i)(x  -  3). 

From  the  above  examples  it  will  be  seen  that  the 
method  of  finding  the  factors  of  a  quadratic  expression 
consists  in  changing  the  expression  into  an  equivalent 
one  which  is  the  difference  of  two  squares. 

The  factors  of  ax^  -\-bx-{-c  are  therefore  found  in  the 
following  manner : 

ax^-{-bx-\-c  =  a[a:^ -\--x-\- -]' 
\        a        aj 

Now  a^H — X  is  made  a  perfect  square,  namely  fx-\ ) , 

•  ^  /"  bY      b^  .    ^       ^^-^ 

by  the  addition  of  (  —  ]  =  — ^-    And,  by  adding  and  sub- 

tracting  — -  to  the  expression  within  brackets,  we  have 
4a^ 

\^        a        a  J         (         a       \2aJ      Aa^     a) 


FACTORS.  145 

The  factors  of  the  expression  in  square  brackets  are, 
by  Art.  115, 

2  a      \\4a2     a)  2a      \\4a^     aj 

Hence  ax^  -\-hx  +  c 

This  formula  solves  the  most  important  case  of  factor- 
ing in  elementary  algebra,  and  in  form  represents  a  class 
of  cases,  also  important,  in  which  the  problem  is:  to 
resolve  an  algebraic  expression  of  higher  degree  in  x,  or 
in  X  and  y,  into  a  product  of  factors  of  the  first  degree 
in  the  same  letters.  In  this  problem  the  coefficients  of 
X  and  of  y  in  the  factors  may,  and  frequently  do,  involve 
irrational  quantities  (surds). 

The  principle  of  factoring,  thus  enunciated,  as  the  stu- 
dent will  have  occasion  to  observe  in  the  sequel,  underlies 
all  elementary  methods  for  solving  algebraic  equations. 

123.  Instead  of  Avorking  out  every  example  from  the 
beginning,  we  might  use  the  formula  found  in  the  last 
article,  and  we  should  only  have  to  substitute  for  a,  h, 
and  c  their  values  in  the  particular  case  we  are  consid- 
ering. 

Ex.  1.   Find  the  factors  of  x^  +  7  a:  -}- 12. 
Here  a  =  1,  &  =  7,  c  =  12. 

Hence         x2  +  7  x  -H  12 

=  {a^+i  +  V(¥-12)}{x  +  |-V(¥-12)} 

=  (x  +  4)(x  +  3). 

K 


146  FACTORS. 

Ex.  2.   Find  the  factors  of  5  a;2  +  32  xy  -  21  y'^. 
Here  a  =  b,  b  =  S2y,  c  =  -21y^. 


=V{ 


266f±my^')  ^19y 
25  /        5  * 


C  5  5Jl  5  6J 

=  6(ix  +  7y)(x-^-^\ 
=  (x  +  7y)(i6x-Sy). 

124.  If  in  the  formula  of  Art.  122  we  put  2  for  a  and 
4  for  6,  we  obtain  the  factors  of  the  expression  a^+2a;4-4 ; 
these  factors  are  x  +  1  +  ^(1  —  4)  and  a;  -f  1  — -^(1  —  4); 
that  is,  a;  +  l4-V(— 3)  and  x-\-l—^{  —  3). 

Now  all  squares,  whether  of  positive  or  of  negative 
quantities,  are  positive.  Hence  it  is  impossible  to  find 
any  real  number  whose  square  is  —3.  Such  an  expression 
as  V— 3  is,  however,  often  used  in  algebra;  and  the 
meaning  to  be  given  to  V— 3  is  simply  that  expressed 
by  V^^  X  V^^  =  -  3. 

The  square  root  of  a  negative  quantity  is  said  to  be 
imaginary  ;  and  a;  + 1  +  V—  3  and  «  + 1  —  V  —  3  are  said 
to  be  imaginary  factors  of  x^  +  2  a;  +  4, 

By  referring  to  the  formula  of  Art.  122,  it  will  be  seen 
that  the  factors  of  ax^  -i-bx-{-  c  are  imaginary  whenever 

• — : is  negative. 

4a^     a 


TACTORS.  14T 

125.  Factors  found  by  Eearrangement  and  Grouping  of 
Terms.  The  factors  of  many  expressions  can  be  found 
by  a  suitable  rearrangement  and  grouping  of  the  terms. 
No  general  rule  can  be  given,  but  in  many  cases  it  will  be 
sufficient  to  arrange  the  expression  according  to  powers  of 
one  of  the  contained  letters,  and  the  factors  will  then  often 
become  obvious,  particularly  if  the  given  expression  only 
contains  one  power  of  that  letter. 

Ex.  1.   Find  the  rational  factors  oix^  -Zx"^ -\- x -Z. 
x8-3a;2  +  x-3  =  x\x  -  3)  +  (x  -  3)  =  (x2  +  i)(a;  -  3). 

Ex.  2.   Find  the  factors  of  ax -\- by  -{■  hx -\-  ay. 
Arrange  in  powers  of  x  ;  then  we  have 

x{a+h)-\-hy  +  ay  =  x{a  +  h)^-y{a+h)  =  (x-\-y){a-\-  b). 

Ex.  3.  Find  the  rational  factors  of  aot^  —  x  -\-  a  —  I. 
Arrange  in  powers  *of  a ;  then  we  have  a  (x^  +  1)  —  (aj  +  1),  and  it 
is  now  obvious  that  ic  +  1  is  a  factor. 

Ex.  4.    Find  the  factors  of  the  first  degree  of 

a\b  -  c)+  fe2  («  -  «)+  c2  (a  -  6). 
Arrange  in  powers  of  a  ;  then  we  have 

a2  (6  -  c)  -  a  (&2  _  c2)  +  ftc  (6  -  c). 
It  is  now  obvious  that  6  —  c  is  a  factor,  and  the  given  expression 
=  (6  -  c){a2  -  a  (&  +  c)+  bc}=(b  -  c)(a  -  6)(a  -  c). 

Ex.  5.   Find  the  factors  of  the  first  degree  of 

a2  _  3  52  _  c2  _  2  a6  +  4  be. 

This  is  a  quadratic  expression  in  a,  or  in  b,  or  in  c.    We  therefore 
proceed  as  in  Art.  122.     The  given  expression 

=  a2  -  2  a6  -  3  62  -  c2  +  4  6c. 


148  FACTORS. 

The  terms  which  contain  a,  namely  a^  —  2  ab,  will  be  made  a 
complete  square  by  the  addition  of  b^.  We  therefore  write  the 
expression  in  the  form 

a2  -  2  a6+  62  _  4  62  _  c2  +  4 6c  =  (a  -  6)2  _  4 62  _  c2  +  4 6c 
=  (a  -  6)2  -(2  6  -  c)2  ={(a  -  6)-(2  6  -  c:)}{(a  -  6)  +  (2  6  -  c)} 

=  (a-36  +  c)(«  +  6-c). 

126.  The  expression  a^  -\-b^-\-  &  —  3abc  is  of  frequent 
occurrence  and  its  factors  should  be  known.  It  is  easy 
to  verify  that 

=  (a  +  6  +  c)  (a^  +  62  _|_  c2  _  5c  -  ca  -  ab). 

The  expression  x'*^  +  x^  + 1  is  also  of  frequent  occur- 
rence. Writing  it  in  the  form  (aj^  +  1)2  —  x^,  we  see  that 
a;^  + 1  4-  07  and  x^  -^1  —x  are  factors  ;  and  we  do  not  pro- 
ceed any  further,  for  both  x"^  -\-x-\-l  and,a^  —  a;  +  1  have 
imaginary  factors  of  the  first  degree. 

The  following  miscellaneous  examples  will  afford  suit- 
able practice  in  the  methods  above  described. 

EXAMPLES  XXXIII. 
Express  in  rational  factors : 

1.  16x4-625y*.  7.  (a  +  6)2-(c-d)2. 

2.  81  a* -16  6*.  8.  (a  +  6)2  -  4  (c  -  ^)2. 

3.  iK5-81a;.  9.  (a  +  6)2  -  (a  +  c)2. 

4.  1  +  27  «».  10.  {X  +  yy  -  (x-  yy. 

5.  27  +  8a;3.  11.  (a  +  6  +  c)2  -  4c2. 

6.  27x*  +  8a;.  12.  (a  +  6  -  3c)2  -  9c2. 

13.  (a2  +  ab  +  62)2  _  (^^2  _  52)2. 

14.  (a2  +  a&  +  62)2 -(a2-a6  + 62)2.     . 

15.  (x3  +  3x)2-(3x2  +  l)2. 


FACTORS.  149 

16.  (a;2  4-  5  ccy  +  y2)2  _  (^^2  -  xy  +  y'^y. 

17.  (a  +  b  +  c  +  d)^-(a-b  +  c-dy. 

18.  (3a  +  26  +  c)2-(a  +  2&  +  3c)2. 

19.  (2x-\-Sy)^-\-(3x  +  2yy.      26.    72  (x2  -  1) -  17 x. 

20.  (a +  36)3 -(3a  +  6)3.  27.    x4-5a;2  +  4. 

21.  ic2  _  3  a;y  _  1  y2.  28.    ic*  -  13a;2?/2  +  36y*. 

22.  x"^  +  la  +  ^\xy  +  y^-  29.  a6  (x2  +  y2)+ a-?/(a2  +  62). 

23.  a;3  +  -y-x2  +  x.  30.  a6(a;2_?/2)+a;?/(a2-62). 

24.  6x'^-5xy-6y^.  31.  (a  +  6)2  -  5c  (a  +  6)+ 6c2. 

25.  X*  -  f  x^y  -  x22/2.  32,  (x  +  y)2  _  7  «  (x  +  ?/)  +  10 z- 

83.    (a  +  6)2  -  8  (a  +  6)(c  +  d!)+  15  (c  +  (Z)2. 
34.  tc(a;  +  2)-y(y  +  2). 

35.  a:  (a; +  4) -?/(?/ +  4).  46.  a262  -  a2  -  6^  +  1. 

36.  x3  -  5  x2  +  a;  —  5.  47.  ac -\-  bd  -  ad  -  be. 

37.  x3  +  ic2  _  4  a;  _  4.  48.  ac2  +  bd^  -  ad^  -  bc\ 

38.  2  a;3  _  3  a;2  _  2  X  +  3.  49.  x^  -  y'^  -\- xz  -  yz. 

39.  5x3-x2-5x+l.  60.  a2  -  6^  -  (a  -  6)2. 

40.  ax8  +  X  +  a  +  1.  51.  a*  +  a262  -  62c2  -  c*. 

41.  ax3  +  6x  +  a  +  6.  62.  a^  -  b'^  +  be  -  ca. 

42.  x3  +  6x2  _  (j2x  _  a25.  53.  a"^  _  a  -  d^  +  c. 

43.  6x3  _|.  ax2  -}-  6x  +  a.  64.  1  +  6x  -  (a2  +  a6)  x2. 

44.  ax2  +  6y2  +  (a  +  6)  xy.  65.  1  -  a6x3  +  (6  -  a2)  x2. 

45.  a262  +  a2  +  62  +  1.  66.  a'^c'^ -{- acd -^  abc  +  bd, 

67.  a2x  +  a6x  -\-  ac  -{-  b'^y  +  aby  +  be. 

68.  a2  -  62  +  c2  -  d^  -  2  (ac  -  6d). 

69.  4  a262  -  (a2  +  62  -  c2)2. 

60.    (a2  -  62  +  c2  -  (|2)2_  (2  ac- 2  6(^)2. 

61.  X4  -  23x2  +  1.  63.    X*  -  11  x22/2  +  y4. 

62.  X*  -  7  x22/2  +  y4.  64.    x4- 3x2+1. 

65.  (x2  +  4x)2  -  2  (x2  +  4x)-  15. 

66.  (x2  +  7  X  +  6)  (x2  +  7  X  +  12)  -  280. 


150  FACTORS. 

127.  Zero  Pactors.  Since  zero  was  defined  [Art.  45]  as 
a  difference,  in  the  form  a  —  a,  a  zero  factor  in  any  prod- 
uct may  be  replaced  by  a  —  a.  If,  then,  F  have  a  finite 
value,  the  product 

0  X  F=  (a  -  a)  F=  aF-  aF 

is  itself  zero,  and  it  is  obvious  that  a  product  will  not 
be  zero  unless  one  of  its  factors  is  zero.  Hence,  exclu- 
sive of  infinite  factors,* 

In  order  that  a  product  may  he  zero,  it  is  necessary  and 
sufficient  that  one  of  its  factors  be  zero. 

128.  Quadratic  Equations.  In  particular,  the  theorem 
of  Art.  127  says :  the  product  {x  —  2){x  —  4)  is  zero,  if 
a;  —  2  is  zero,  or  if  aj  —  4  is  zero  ;  and,  in  order  that  this 
product  may  be  zero,  one  of  these  factors  must  be  zero. 

The  roots  of  the  equation 

(a;  -  2)  (a;  -  4)  =  0 

are  therefore  2  and  4.     [Definition  of  a  root,  Art.  91.] 
Again,  the  product 

(a; -3)  (a;- 4)  (a; -5) 
vanishes   if  a;  —  3  =  0,  or  if  a;  —  4  =  0,  or  if  aj  —  5  =  0, 

*  The  combination  of  zero  and  infinite  factors  in  a  product  introduces 
what  are  known  as  indeterminate  forms.  Thus,  for  the  value  x  =  a, 
the  expression  (x'^  — a^)x  1/ (x  —  a)  assumes  the  form  0x1/0,  or 
0  X  00 ,  where  1  /  0  is  said  to  be  infinite  and  is  represented  by  the  symbol 
00.  But  this  product  is  not  necessarily  zero,  when  x  =  a;  for,  suppress- 
ing the  common  factor  x  —  a  from  x^  —  a^,  and  x  —  a,  we  have 

(x2  -  a2)  X  1  /  (x  -  a)  =  ^^~^^  =  X  +  a , 
x  —  a 

and  X  +  a  =  2  a  when  x  =  a.    [See  Treatise  on  Algebra,  Art.  217.] 


FACTORS.  161 

and  in  no  other  case ;  therefore  3,  4,  and  5  are  the  roots 
of  the  equation 

{x  -  3)  (a;  -  4)  (cc  -  5)  =  0. 
Applied  to  the  general  quadratic  expression  ax^  -\-hx  +c, 
this  principle  of  vanishing  products  asserts  that 

ax'^  -\-hx-\-Cy 
or  its  equivalent  [Art.  122] 

«{-f.W(&-9}{-^-V{£-;)} 

becomes  zero,  if  and  only  if  one  of  the  factors  here  writ- 
ten is  zero,  and  that  therefore  the  values  of  x  that  satisfy 
the  equation 

aa^  -\-hx-\-c  =  0 

are  derived,  one  of  them  from  the  equation 


x  + 


2^^Vfe~a)"^' 


the  other  from  the  equation 

Thus  the  problem  of  solving  any  equation  of  the 
second  degree  is  reduced  to  that  of  solving  two  equations 
of  the  first  degree.  This  subject  will  be  taken  up  again 
in  the  chapter  on  quadratic  equations. 

The  general  principle,  of  which  the  above  examples 
are  particular  cases,  asserts  that  the  equation 

{x  -a){x-  h)  {x  -  c) =  0 

is  equivalent  to  the  system  of  alternative  equations 
a;  —  a  ==  0,  or  x  —  h  =  0,  or  a;  —  c  =  0,  etc. 

Note.  —  Observe  the  distinction  we  make  between  alternative 
equations  and  simultaneous  equations  [Art.  101] . 


152  FACTORS. 

129.  From  the  examples  considered  in  the  last  article 
it  will  be  apparent  that  the  solution  of  an  equation  of  any 
degree  can  he  written  down  at  once  provided  the  equation  is 
given  in  the  form  of  a  product  of  factors  of  the  first  degree 
equated  to  zero. 

The  following  are  examples  of  such  equations : 

Ex.1.     Solve         (x-l)(x+l)  =  0. 

The  equation  is  satisfied  if  ic  —  1  =  0,  or  if  x  +  1  =  0,  and  in 
no  other  case. 

Hence  we  must  have 

x-l=0,OTx-]-l  =  0; 
that  is,  a;  =  1,  or  X  =  —  1. 

Thus  the  roots  of  the  equation  are  1  and  —  1. 

Ex.2.     Solve      x(x+l){x  +  2)=0. 

The  equation  is  satisfied  if  ic  =  0,  or  if  a;  +  1  =  0,  or  if  a;  +  2  =  0, 
and  in  no  other  case. 

Hence  we  must  have 

x  =  0,  ora;  +  l  =  0,  orx  +  2  =  0; 
that  is,  a:  =  0,  or  oj  =  —  1,  or  a;  =  —  2. 

Thus  the  roots  of  the  equation  are  0,-1,  and  —  2. 

Ex.3.     Solve        x(2a;-l)(2a;  +  3)  =  0. 

The  equation  is  satisfied  if  cc  =  0,  or  if  2  a;  —  1  =  0,  or  if 
2  X  +  3  =  0,  and  in  no  other  case. 

Hence  we  must  have 

x  =  0,  or2x-l  =  0,  or2x  +  3  =  0; 
that  is,  X  =  0,  or  X  =  |,  or  x  =  -  f . 


130.    Since  all  the  terms  of  any  equation  can  be  trans- 
posed to  one  side,  an  equation  can  always  be  written  with 


FACTORS.  153 

all  its  terms  on  one  side  of  the  sign  of  equality,  and  zero 
on  the  other  side. 

It  therefore  follows  from  the  last  article  that  tte  prob- 
lem of  solving  an  equation  of  any  degree  is  the  same  as  the 
problem  of  finding  the  factors  of  an  expression  of  the  same 
degree. 

Hence  the  process  of  solving  any  equation  which  in- 
volves only  one  unknown  quantity  is  as  follows : 

First  write  the  equation  with  all  its  terms  on  one  side  of 
the  sign  of  eqvMity,  and  zero  on  the  other  side ;  then  resolve 
the  whole  expression  into  factors  {of  the  first  degree  in  the 
unknown  quantity),  and  the  values  obtained  by  equating 
each  of  these  factors  separately  to  zero  will  be  the  required 
roots. 

In  the  following  examples  the  resolution  into  factors 
can  be  performed  by  inspection. 

Ex.  1.     Solve  the  equation  x^  —  3  x  =  0. 

Since  x^  —  3  x  =  x  (x  —  3), 

we  have  x  (x  —  3)  =  0. 

Whence  x  =  0,  or  x  —  3  =  0 ; 

the  roots  required  are  therefore  0  and  3. 

Ex.  2.     Solve  the  equation  x^  —  9  =  0. 

Since  x2  -  9  =  (x  -  3)  (x  +  3), 

we  have  (x  -  3)  (x  +  3)  =  0. 

Whence  x-3  =  0,  orx  +  3  =  0; 

the  roots  required  are  therefore  3  and  —  3. 

Ex.  3.     Solve  the  equation  x^  —  2  =  0. 

Since  x2  -  2  =  (x  -  v2)  {oc  +  y/2), 

we  have         (x  -  ^2)  (x  +  ^2)  =  0. 
Hence  x  —  y^  =  0,  or  x  +  v^  =  0. 

Thus  y/2  and  —  ^2  are  the  required  roots  of  the  equation. 


154 


FACTORS. 


Ex.  4.     Solve  the  equation  x^  -  4x  =  0. 

Since  x^  -  4x  =  x  (x^  -  4)  =  x  (x  -  2)(x  +  2)^ 

-  we  have  x(x  -  2)(x  -{-  2)  =  0. 

Hence  x  =  0,  or  a:  -  2  =  0,  ot  x  +  2  =0. 

Thus  0,  2,  and  —  2  are  the  roots  of  the  equation.   ■ 

Ex.  5.     Solve  9x^  =  ix. 

Transposing,  we  have 

9x^-4:x  =  0; 
that  is,  x(9x^  -  i)  =  0, 

or  x(Sx-2)(Sx-{-2)  =  0. 

Hence  a:  =  0,  or  .3  a;  -  2  =  0,  or  3  x  +  2  =  0  ; 

that  is,  X  =  0,  or  a;  =  I,  or  X  =  -  |. 

Ex.  6.    Solve  x2  +  6  =  5  X. 

Transposing,  we  have 

x2-5x  +  6=0; 
that  is,  (X  -  2)  (x  -  3)  =  0. 

Hence  x  -  2  =  0,  or  x  -  3  =  0. 

Thus  2  and  3  are  the  required  roots. 

EXAMPLES  XXXIV. 
Solve  the  equations 

1.  (x-l)(x  +  2)  =  0.  7.   x(x-2)(x  +  3)  =  0. 

2.  (x-3)(x-4)=0.  8.   x(x  +  2)(x-4)  =  0. 

3.  (x  +  l)(x  +  2)  =  0.  9.   x(x-3)(x  +  4)=0. 

4.  (2x  +  l)(2x-l)=0.  10.   x(2x-l)(3x  +  4)  =  0. 

5.  (3x-l)(3x  +  l)  =  0.  11.  x(5x-2)(6x-7)  =  0. 

6.  (2x-5)(x-4)=0.  12.   x(x-3)(3x+7)  =  0. 

13.  (x-l)(x-2)(x-3)(x-5)  =  0. 

14.  (x-2)(x-l)(x+l)(x  +  2)  =  0. 

15.  (3x-l)(4x+ l)(5x-2)(2x  +  7)  =  0. 

16.  (2x-3)(3x-4)(4x-6)(5x  +  6)=0. 


FACTOKS.  155 

17.  x^-x  =  0.  25.  3x2  =  6a;. 

18.  x^-2x  =  Q.  26.  5ic2  =  -6x. 

19.  x2  +  3x  =  0.  27.  x2  =  6x. 

20.  2  x2  -  3  X  =  0.  28.  ax2  =  6x. 

21.  2x2-5x  =  0.  29.  6x2  =  x-2  +  5. 

22.  3x2  +  x  =  0.  30.  5(x2  +  5)=3(x2  +  25). 

23.  x2  =  6x.  31.  5(x2  +  4)=4(x2  +  9). 

24.  2x2  =  x.  32.  2(x2  +  7)=7(x2  +  2). 

33.  5(x2  +  3)-(x-5)(x  +  5)=76. 

34.  7(x2-l)-(x  +  3)(x-3)  =  56. 

35.  3x2  +  (5x  +  2)2  =  20x  +  32.  43.   x2-6x-84  =  0. 

36.  17  +  3  X  =  |(x  +  3)2  _  28.  44.    x2  -  x  -  156  =  0. 

37.  x2-5x  +  6  =  0.  45.   x2+5x- 150  =  0. 

38.  x2-7x+12  =  0.  46.   x2  +  2x  =  3. 

39.  x2-12x  +  20  =  0.  47.    x2  +  4x  =  45. 

40.  x2  -  9x  +  20  =  0.  48.   x2  -  3 x  =  10. 

41.  x2  -  11  X  +  28  =  0.  49.    x3  +  11  x2  -  180  x  =  0. 

42.  x2-25x  + 150  =  0.  60.  x8-x2  =  132x. 


156  HIGHEST   COMMON   FACTORS. 


CHAPTER  XI. 

Highest  Common  Factors. 

131.  A  Common  Factor  of  two  or  more  algebraic  expres- 
sions is  an  expression  which  will  exactly  divide  each  of 
them. 

Thus  a  is  a  common  factor  of  ab  and  ac. 

The  Highest  Common  Factor  of  two  or  more  algebraic 
expressions  is  the  expression  of  highest  dimensions  which 
will  exactly  divide  each  of  them. 

Thus  a^  is  the  highest  common  factor  of  a^b  and  a^c. 

Instead  of  Highest  Common  Factor  it  is  usual  to  write 
H.C.F. 

We  proceed  to  show  how  to  find  the  H.  C.  F.  of  given 
expressions. 

132.  H.  C.  F.  of  Monomial  Expressions.  The  highest  com- 
mon factor  of  two  or  more  monomial  expressions  can  be 
seen  by  inspection. 

Take,  for  example,  a^'^c  and  a%^(^, 

The  first  expression  is  clearly  divisible  by  a,  or  by  a^,  or  by  a^, 
but  by  no  higher  power  of  a  ;  and  the  second  expression  is  divisi- 
ble by  a,  or  by  a'^,  but  by  no  higher  power.  Hence  a^  will  divide 
both  expressions,  and  it  is  the  highest  power  of  a  which  will  divide 
them  both.  Also  b'^,  but  no  higher  power  of  b,  will  divide  both 
expressions ;  and  c,  but  no  power  of  c  above  the  first,  will  divide 
both  expressions. 


HIGHEST  COMMON   FACTORS.  167 

Hence  the  H.C.F.  of  a^h^-c  and  a^ftV  is  a^y^c. 

Again,  to  find  the  H.  C.  F.  of  ahHH"^  and  aH^d^. 

The  highest  power  of  a  which  divides  both  expressions  is  a ;  6 
will  not  divide  both  expressions ;  the  highest  power  of  c  which 
divides  both  is  c^  ;  and  the  highest  power  of  d  which  divides  both 
is  (?*.     Hence  the  H.  C.  F.  is  acH\ 

Also  a^hd^^  a%HH^  and  a^hd^d^  are  all  divisible  by  a^,  by  6,  and 
by  c*  ;  and  therefore  the  H.  C.  F.  of  the  three  expressions  is  a%c^. 

From  the  above  examples  it  will  be  seen  that  the 
H.C.F.  of  two  or  more  monomial  expressions  is  obtained 
by  taking  each  letter  which  is  common  to  all  the  expres- 
sions to  the  lowest  power  in  vjhich  it  occurs. 

If  the  expressions  have  numerical  coefficients,  the 
G.  C.  M.  of  these  can  be  found  by  arithmetic,  and  pre- 
fixed as  a  coefficient  to  the  algebraical  H.  C.  F. 

EXAMPLES  XXXV. 
Find  the  H.  C.  F.  of 

1.  a862anda268.  7.  9  a^ftSa^y^  and  8  xV- 

2.  a6c2  and  a25c8.  g.   f  a^ft^cs  and  2  ft^c*. 

8.  9  a68  and  6  a%.  9.  42  axy"^^  and  77  hhf^. 

4.  4  x^y  and  10  xy^.  10.  ah^,  a^bc,  and  abc^. 

6.  2i  a%^x^  a.nd  eO  a'^b*3fi.  11.  Sx'2yz^,l5xy^z'^,Sindl0x'^y^z'^. 

6.  a'^b^x^  and  3  b^x.  12.  ab^c^x\  a^bc^x^,  and  a^b'^cx^. 

133.  H.  0.  F.  of  Multinomial  Expressions  whose  factors  are 
known.  When  the  factors  of  two  or  more  multinomial 
expressions  are  known,  their  H.  C.  F.  can  be  at  once 
written  down.  The  H.  C.  F.  will  be  the  product  obtained 
by  taking  each  factor  which  is  common  to  all  the  expres- 
sions to  the  lowest  power  in  which  it  occurs. 


158  HIGHEST   COMMON   FACTORS. 

Consider,  for  example,  the  expressions  (x  —  ay(x+  b)^  and 
(x  —  ay{x  +  by.  It  is  clear  that  both  expressions  are  divisible  by 
(x  —  ay,  but  by  no  higher  power  of  (x—  a).  Also  both  expressions 
are  divisible  by  (x  +  by,  but  by  no  higher  power  of  x  +  b. 

Hence  the  H.  C.  F.  is  (x  -  ay(x  +  bj^ 

Again,  a^b  (a  -  b)'^  {a  +  by  and  ab^  (a  -  by  (a  +  by  are  both 
divisible  by  a,  by  6,  by  («  —  by,  and  by  (a  +  by.  Hence  the 
H.  C.  F.  is  ab  (a  -  by  (a  +  by. 

In  the  following  examples  the  factors  can  be  seen  by- 
inspection,  and  hence  the  H.  C.  F.  can  be  written  down. 

Ex.  1.     Find  the  H.  C.  F.  of  a%^  -  a%^  and  a^&a  +  a%^. 

Since    a^b"^  -  a^b^  =  a^b^  (a^  -  6^)  :^  0^2^,2  (^^  _i)^(^a  +  b), 
and  a453  4.  ^354  ^  ^3^3  (^  +  6), 

we  see  that  the  H.  C.  F.  is  a'^b'^  (a  +  b). 

Ex.  2.     Find  the  H.  C.  F.  of 

«3  +  3  a'^b  +  2  a&2  and  a*  +  4  a^ft  +  3  aS^a. 
a^  +  Sa'^b  +  2  ab'^  =  a  (a^  +  3  a&  +  2  6-2)  =  a  (a  +  &)  (a  +  2  6), 
and 
a*  +  4  a35  +  3  ^2^2  =  cfi  ((^2  +  4^5  +  3  52)  =a'^(^a+b){a  +  ^b)\ 

hence  the  H.  C.  F.  is  a  (a  -f  b). 

Ex.  3.  Find  the  H.  C.  F.  of  3  a^  +  2  a2  _  «  and  5  a*  +  3  a^  _  2  a2, 

3  a3  +  2  a2  -  a  :=  a  (3  a2  +  2  a  -  1)  =  a  (3  a  -  1)  (a  +  1), 
and    5  a*  +  3  a3  _  2  a2  ^  ^2  (5  ^2  +  3  q^  _  2)  =  ^2  (5  ^^  _  2)  (a  +  1)  ; 

hence  the  H.  C  .F.  is  a  (a  +  !)• 

EXAMPLES  XXXVI. 
Find  the  H.  C.  F.  of 

1.   x2  (x  -  ay  and  x^  {x  -  ay.  2.    a2  _  52  and  (a  +  by. 

3.  3  a6  (a  +  6)2  and  2  (a  -  &)  (a  +  &)3. 

4.  64^6  (5  _i_  c)2  and  ft^cS  (&  +  cy. 

6.    ^353  ^_  ab^  and  a^  _  <x254.        7.    (^-2^3  +  2  a3x2  and  aH'^  -  4  ^4x2. 
6.   a26  +  3  63  and  a^  -  9  a'^b^.      8.    a''x2  -  4  a+x^  and  a6x2  _  16  a'^oi^. 


HIGHEST   COMMON   FACTORS. 


159 


9.  x2  +  3  X  +  2  and  x^  +  6  x  +  8. 

10.  x3  +  3  x2?/  +  2  X2/2  and  X*  +  6  o(?y  +  8  xV- 

11.  3x2-4x+ 1  and4x2-5x+ 1. 

12.  3  a2  -  4  a6  +  62  and  4  a*  -  5  a^h  +  a262. 

v/  134.  Although  we  cannot,  in  general,  find  the  factors 
'oi  a  multinomial  expression  of  higher  degree  than  the 
second,  we  can  always  find  the  highest  common  factor  of 
any  two  expressions  by  the  following  process,  which  will 
be  seen  to  be  analogous  to  that  used  in  arithmetic  to  find 
the  greatest  common  measure  of  two  numbers. 

Eule.  Arrange  the  two  expressions  in  descending  powers 
of  some  common  letter,  and  divide  the  expression  ivhich  is 
of  the  highest  degree  in  the  common  letter  by  the  other :  if 
both  expressions  are  of  the  same  degree,  it  is  immaterial 
which  is  used  as  the  divisor.  Take  the  remainder,  if  any, 
after  the  first  division  for  a  new  divisor,  and  the  former 
divisor  as  dividend;  and  continue  the  process  until  there 
is  no  remainder.  The  last  divisor  will  be  the  H.  C.  F.  of 
the  two  given  expressions. 

For  example,  to  find  the  H.  C.  F.  ofx^  +  x2  -  2  and  x^  +  2  x2  -  3, 
the  process  is  as  follows  : 


First  dividend, 


Second  dividend, 


Third  dividend, 


Thus  the  H.  C.  F.  is  x  -  1. 


x8  + 

X3- 

a;8  4.  2  x2  -  3 

X8  -1-       X2  -  2 

x2-2 

X 

X2-1 
X    +1 

x^-\-x  ^2 

X2              -1 

•    • 

X2-1 

X    -1 

X2-X 

X    +1 

X    -1 
X    -1 

X*  +  x2  —  2    First  divisor 


Second  divisor 


Third  divisor 


160 


HIGHEST  COMMOIT  FACTORS. 


The  work  can  be  arranged  more  concisely  as  follows : 


a:3+    3^2  _  2 

x^  +  x^  -2 
x^-x 

x^-1 

x'^-x 

jc2  4-  X  -  2 

X2               -1 

X  -1 
X    -1 

a;  -1 

Note.  —  It  is  only  multinomial  factors  which  are  to  be  found  by 
the  above  rule  ;  the  H.  C.  F.  of  any  monomial  factors  of  the  given 
expressions  must  be  found  by  inspection. 


For  example,  to  find  the  H.  C.  F.  of 

a2^4  _^  Qj2^3  _  2  «2  ^  and  abx^  +  2  abx* 
Since  ^2.y.i  _l  «2^3 

and 


-  3  abx^. 
a'x'  +  a^x""  —  2aH  —  aH  (x^  +  a^^  -  2), 
abx^  +  2  «&x*  -  3  a6x2  _  ^^5^2  (3^3  +  2  a;2  _  3), 


it  is  clear  that  the  H.  C.  F.  of  the  monomial  factors  is  ax.  We  have 
already  found  that  the  H.  C.  F.  of  x^  +  a;2  _  2  and  x^  +  2  x2  -  3 
is  X  —  1. 

Hence  the  whole  H.  C.  F.  is  ax  (x  —  1) . 

'  )^35.  The  rule  given  in  Art.  134  for  finding  the  H.  C.  F. 
of  two  expressions  which,  have  no  monomial  factors,  may 
be  proved  as  follows. 

Let  A  and  B  stand  for  the  two  expressions,  which  are 
supposed  to  be  arranged  according  to  descending  powers 
of  some  common  letter,  and  let  A  be  of  not  higher  dimen- 
sions than  B  in  the  common  letter. 

Let  B  be  divided  by  A^  and  let  the  quotient  be  Q,  and 

It  the  remainder,  so  that  the  first  stage  of  the  process  of 

finding  the  H.  C.  F.  of  A  and  B  is  as  under, 

A)B{q 

AQ 

B 


HIGHEST  COMMON   FACTORS.  161 

Then  we  have  B  =  AQ  +  R (i.), 

and  E  =  B-AQ (ii.). 

Now  an  expression  is  exactly  divisible  by  any  other  if 
each  of  its  terms  is  so  divisible;  hence,  from  (i,),  B  is 
divisible  by  any  common  factor  of  A  and  R. 

Thns  any  common  factor  of  A  and  Ii  is  also  a  common 
factor  of  A  and  B. 

Again,  any  factor  of  both  B  and  A  will  be  a  factor  of 
B  —  AQ',  that  is,  from  (ii.),  will  be  a  factor  of  M. 

Thus  any  common  factor  of  A  and  B  is  also  a  common 
factor  of  A  and  R. 

It  follows  therefore  that  the  common  factors  of  A  and 
B  are  exactly  the  same  as  the  common  factors  of  A  and  R. 
Hence  the  H.  C.  F.  of  A  and  B  is  the  H.  C.  F.  of  A  and  R. 

If  now  we  divide  A  by  R,  and  the  remainder  is  S,  the 
H.  C.  F.  of  S  and  R  will  similarly  be  the  same  as  the 
H.  C.  F.  of  A  and  R,  and  therefore  will  be  the  H.  C.  F. 
required.  And  so  on ;  so  that  the  H.  C.  F.  of  any  divisor 
and  the  corresponding  dividend  is  the  H.  C.  F.  required. 

But  if  at  any  stage  there  is  no  remainder,  the  divisor 
must  be  a  factor  of  the  corresponding  dividend,  and  that 
divisor  is  clearly  the  H.  C.  F.  of  itself  and  the  corre- 
sponding dividend.  It  must  therefore  be  the  H.  C.  F. 
required. 

It  should  be  remarked  that  by  the  nature  of  division 
the  remainders  are  successively  of  lower  and  lower  dimen- 
sions ;  and  hence,  iHiless  the  division  leaves  no  remainder 
at  some  stage,  we  must  at  last  come  to  a  remainder  which 
does  not  contain  the  common  letter,  in  which  case  the 
given  expressions  have  no  H.  C.  F.  containing  that  letter. 

We  have  already  remarked  that  the  process  we  are 


162 


HIGHEST  COMMON   FACTORS. 


considering  is  only  to  be  used  to  find  the  H.  C.  F.  of  the 
multinomial  factors  of  the  given  expressions ;  and,  bear- 
ing this  in  mind,  it  is  clear  that  any  of  the  expressions 
which  occur  may  be  divided  or  multiplied  by  any  mono- 
mial expression  without  destroying  the  validity  of  the 
process ;  for  the  multinomial  factors  will  not  be  altered 
by  such  division  or  multiplication. 

Ex.1.    Find  the  H.  C.  F.  of 

x^-\-4:X^  -Sx  +  24:  and  x^  -x^  +  8x-8. 
Neither  expression  has  any  monomial  factors;   we  therefore 
proceed  as  in  Art.   134,  performing  the  work  by  the  synthetic 
method. 


X3 

X*- 

-x3+0      +8a:-8 

-4x2 

- 

-4x3  +  20x2 

+  8x 

+    8  x2  -  40  X 

-24 

-  24  X  +  120 

x-5;         28x2 -56x  + 112 
The  remainder  23  x2  -  56  x  -f  112  =  28  (x2  -  2  x  +  4)  ;  and  since 
the  numerical  factor  28  is  clearly  not  a  factor  of  the  given  expres- 
sions, we  reject  it,  and  continue  the  process  with  x2  —  2  x  +  4  as 
the  new  divisor : 


+  2x 
-4 


x3  +  4x2-    8X-1-24 
+  2x2  +  12x 

-    4x-24 


x  +  6;  0 

Thus  the  H.  C.  F.  is  x2  -  2  x  +  4. 


0 


Ex.  2.   Find  the  H.C.F.  of  x^-4a^x+lba^  and  x*+a2x2+25a*. 


X3 

+  4a2x 
-15a2 


x*  +  0  +    a2^2  +  0 

+  4  «2x2 

—  15  a% 


+  25a4 


x;  5a2x2_  I5a3x  +  25a4 

Now      5  a2x2  -  15  a^x  +  25 a*  =  5  a2(x2  -  3  ax  +  5  a^). 


HIGHEST   COMMON   FACTORS. 


163 


We  reject  the  monomial  factor  5a^,  and  continue  the  process 
with  x2  -  3  ax  +  5  a-  as  the  new  divisor : 

x3  +  0       -  4  a2x  +  15  a3 
+  3ax2+9a2x 

-  5  a2x  -  15  a3 


x^ 
+  3ax 
-5a2 


x  +  3a;      0  0 

Hence  x^  —  3  ax  +  5  a^  is  the  required  H.  C.  F. 

Ex.  3.   Find  the  H.  C.F.otx^-y^  and  x'  -  y"^. 

x'  —  y"^    |x^  —  y^ 


xV|x2 


Before  using  the  remainder  as  a  divisor,  we  reject  the  monomial 
factor  y^ ;  we  then  have  x^  _  yi. 


x6-y5 

x2-!/2 

X6  -  x82/2 

X8  +  X«/2 

X«y2 

xV 

_y6 
-Xt 

xy*  -  y5 
We  reject  the  monomial  factor  y*  of  the  remainder. 


xy 


x  +  y 


xy 
xy_ 


t. 


Hence  x  —  y  is  the  required  H.  C.  F. 

Ex.  4.  Find  the  H.  C.  F.  of  2x2-5x  +  2  and  x3+4x2-4x-16. 
To  avoid  the  inconvenience  of  fractions,  we  multiply 
x8  +  4x2-4x-16 

by  2.  We  have  shown  in  the  preceding  article  that  we  may  do 
this,  as  no  additional  common  factors  can  be  thereby  introduced. 
It  will  be  seen  that  we  multiply  a  second  time  in  the  course  of  the 
work.    The  following  arrangement  will  be  found  convenient : 


164 


HIGHEST   COMMON   FACTORS. 


First  dividend  x^  +  4:X^-    4  x  -  16 

Multiply  by     .     .    2    First  divisor. 


X 

2x3  +  8^2-    8a: -32 
2  a:3  -  5  x2  +    2  x 

2x2- 5a: +  2 

2x2 -4a: 

13x'^-  lOx-32 
Multiply  by     .     .    2 

-x  +  2 
-a:+2 

13 

26a:2_20a:-64 

26x2 -65  a: +  26 

0      0 

2  a: 


Divide  by     .     . 
Second  divisor . 


45i45x-90 
X-    2 


Thus  a:  -  2  is  the  H.  C.  F. 

136.  The  H.  C.  F.  of  more  than  two  expressions  is 
sometimes  required,  when  the  factors  cannot  be  deter- 
mined by  inspection. 

Now  it  is  clear  that  any  factor  which  is  common  to 
each  of  t'hree  or  more  expressions,  must  be  a  factor  of 
the  H.  C.  F.  of  any  two  of  them.  Hence  we  first  find  the 
H.  C.  F.  of  two  of  the  given  expressions,  and  then  find 
the  H.  C.  F.  of  this  result  and  of  the  third  expression, 
and  so  on. 


Ex.     Find  the  H.C.F.  of 

x3  +  x2-x-l,  a:3  +  3a:2 


X 


3,  and  a:3  +  x2-2. 


The  H.  C.  F.  of  the  first  two  expressions  is  a:2  -  1.  The  H.  C.F, 
of  a:2  —  1  and  the  third  of  the  given  expressions  is  x  —  1.  Hence 
X  —  1  is  the  H.  C.  F.  required. 


137.  If  the  numerical  coefficients  of  the  first  terms  of 
the  given  expressions  are  large,  it  is  often  desirable  to 
arrange  the  terms  of  both  expressions  in  the  reverse 
order  before  applying  the  rule  for  finding  their  H.  C.  F. 


HIGHEST   COMMON    FACTORS.  165 

For  example,  in  order  to  find  the  H,  C.  F.  of 

7  x*  +  2x3  -  a:-2  +  8x  +  1  and  9x*  -  2  a;3  +  3  a;2  -f  6  X  +  1 
it  is  best  to  arrange  the  expressions  in  the  form 

1  +  8x-x2  +  2x3  +  Tx^andl  +6x  +  3x2  -2x3  +  9x*. 

138.  The  highest  common  factor  of  two  expressions  is 
sometimes,  but  very  inappropriately,  called  their  greatest 
common  measure  [G.  C.  M.]. 

The  inappropriateness  of  the  term  greatest  common 
measure  can  be  seen  by  an  example.  If  a^  is  a  factor  of 
two  expressions,  so  also  is  a,  and  a^  is  of  higher  dimen- 
sions than  a ;  but,  as  a  may  represent  any  number  what- 
ever, a^  is  not  necessarily  greater  than  a ;  in  fact,  if  a  is 
positive  and  less  than  unity,  d^  is  less  than  a. 

It  should  also  be  noticed  that  if  we  give  particular 
numerical  values  to  the  letters  involved  in  any  two  ex- 
pressions, and  in  their  H.  C.  F.,  the  numerical  value  of 
the  H.  C.  F.  is  by  no  means  necessarily  the  G.  C.  M.  of  the 
values  of  the  expressions.  This  may  not  be  the  case  even 
when  the  given  expressions  are  integral  for  the  particular 
values  chosen.  For  example,  the  H.C.F.  of  2a^-f-15a;-f  13 
and  6  a;^  -t- 17  a;  +  11  will  be  found  to  be  a;  -f  1 ;  but  if  we 
suppose  X  to  be  \^  the  numerical  values  of  both  expres- 
sions, and  therefore  their  G.  C.  M.,  will  be  21,  whereas 
the  numerical  value  of  the  H.  C.  F.  will  be  f . 

EXAMPLES   XXXVII. 
Find  the  H.C.F.  of 

1.  x2  -  6  X  -I-  4  and  x^  -  5  x^  +  4. 

2.  x2  —  5  xy  -|-  4  y2  and  x*  —  5  x^y  +  4  xy^. 

3.  2x2 -6x  + 2  and  4x3+ 12x2- X- 3. 


166  HIGHEST   COMMON   FACTORS. 

4.  2 x2  -  5 a;?/  +  2  y2  and  4^:^  +  12 x^^y  -  xy^  -  S yK 

5.  a;*  +  3  x2  -  10  and  a:4  -  3  0^2  +  2. 

6.  x^  +  3x*y  -  10 x2?/2  and  x^-Sx^y  +  2 y^. 

7.  2 a2  -  5 a  +  2  and  2  a3  -  3a2  _  8a  +  12. 

8.  2  62  _  5  5  +  2  and  12  63  _  8  6-2  _  3  5  _^  2. 

9.  ic2  -  3x  +  2  and  x^  -Sx-j-  2. 

10.  xY  -  3  a;2^2  +  2  and  x^y^  -  S  x^y^  +  2. 

11.  x^-Sa^x-2 a^  and  x^  -  aic2  _  4 a^. 

12.  2  a3  +  3  a26  -  6^  and  4  a^  +  ^62  _  ^,3. 

13.  a^  +  68  and  a*  +  a262  +  54. 

14.  8  a3  +  1  and  16  a*  +  4  a2  +  1. 

15.  x3  +  2  ic2y  -  a;^/2  _  2  ?/3  and  x^  -  2  a:2y  -  xy'^  +  2  y^. 

16.  2x3  +  5x2  +  X  -  8  and  3x3  -  4x2  4-  9x  -  8. 

17.  3x3  +  x2  +  X  - 2  and  2x3  -  x2  -  x  -  3. 

18.  x3  -  4x  +  15  and -X*  +  x2  +  25. 

19.  3  x2  -  38x  +  119  and  x3  -  19x2  +  119x  -  245. 

20.  3x3  -  3x2y  +  X2/2  -  2/3  and  4x2?/  -6xy^  +  y^. 

21.  12x2-  15x?/  +  32/2  and  6  x3  -  6  x2?/ +  2  x?/2  -2y^. 

22.  2  x2  -  14  X  +  20  and  4x  (x2  +  5)  -  25  (x  +  l)(x  -  1). 

23.  16  X*  +  4  x2  +  1  and  8  x*  -  16  x3  +  x  -  2. 

24.  a2_4a;2+ 12X-9  and  a2  + 2a-  '  8x-3. 

25.  21  -  7  X  +  3x2  -  x3  and  35  +  19  x2 

26.  2x*  +9x3  +  14x  +  3  and  2  +  9x  +  14x3 +  3x*. 

27.  x3  +  3  x2  -  X  -  3  and  x*  +  4  x3  -  12  x  -  9. 

28.  3x* +  5x3-7x2  +  2x4-2  and  2x*  +  3x3 -2x2  +  12x  +  5. 

29.  y3  _  2?/2  +  3  2/  _  6  and  2/4  -  2/3  _  2/2  _  2  y. 


HIGHEST  COMMON   FACTORS.  167 

30.  y*  -  3  yz^  +  20  z^  and  5y^  -Sy^z-^  64  z^. 

31.  2ic3_iix2  4-  llx+4  and  2x4-3x3  +  7x2-  12x-4. 

32.  2x4  +  4x3  +  3x2-2x-2  and  3x*  +  6x3  +  7x2  + 2a +  2. 

33.  x*  -  x3  +  2  x2  -  X  -  1  and  2  X*  -  2  x2  +  X  -  1. 

34.  2x*-6x3  +  3x2-3x+  1  and  x^  -  3x^  +  x«  -  4x2  +  12x-4. 
86.  x8  +  (to  -  3)  x2  -  m  (2m  +  3)  X  +  6 m^ 

and  x8  +  (5TO-3)x2  +  3m(2m-  5)x-18to2. 
86.   mn  (x^  +  y^)  -\-xy  (m^  +  ri^)  and  mn  (x^  +  y^)  +xy  (m^y  +  w^x). 


^•c* 


168  LOWEST   COMMON   MULTIPLES. 


CHAPTER  XII. 

Lowest  Common  Multiples. 

139.  A  Common  Multiple  of  two  or  more  algebraical 
expressions  is  an  expression  whicli  is  exactly  divisible 
by  each  of  them. 

The  Lowest  Common  Multiple  of  two  or  more  algebraical 
expressions  is  the  expression  of  lowest  dimensions  which 
is  exactly  divisible  by  each  of  them. 

Instead  of  lowest  common  multiple  it  is  usual  to  write 
L.  C.  M. 

We  proceed  to  show  how  to  find  the  L.  C.  M.  of  given 
expressions. 

140.  We  first  consider  monomial  expressions,  the 
factors  of  which  can  be  seen  by  inspection. 

Take,  for  example,  a%'^c  and  a%H^.  The  highest  power  of  a 
which  occurs  in  either  expression  is  a^ ;  hence  any  common  multi- 
ple of  the  given  expressions  must  contain  a^  as  a  factor.  Any- 
common  multiple  must  also  contain  h^  as  a  factor,  and  it  must 
also  contain  c*  as  a  factor.  Any  common  multiple  must  therefore 
contain  a^^C^  as  a  factor ;  and  hence  the  common  multiple  of 
lowest  dimensions  must  be  a^6^c*. 

Again,  to  find  the  L.  C.  M.  of  a%c^,  a%^cH  and  a'^hcM'^,  the 
highest  power  of  a  in  the"  given  expressions  is  a* ;  hence  any  com- 
mon multiple  of  them  must  contain  a*  as  a  factor.  Any  common 
multiple  must  also  contain  6^^  ^s^  ^^(i  ^-2  ^g  factors.  Any  common 
multiple  must  therefore  contain  d^hH^d'^  as  a  factor;  hence  the 
L.  C.  M.  required  is  a^c^hH^. 


LOWEST   COMMON  MULTIPLES.  169 

From  the  above  examples  it  will  be  seen  that  the 
L.  C.  M.  of  two  or  more  simple  expressions  is  obtained 
by  taking  every  letter  which  occurs  in  the  different^  expres- 
sions to  the  highest  power  which  it  has  in  any  one  of  them. 

If  the  expressions  have  numerical  coefficients,  the 
L.  C.  M.  of  these  can  be  found  by  arithmetic,  and  pre- 
fixed as  a  coefficient  to  the  algebraical  L.  C.  M. 

EXAMPLES  XXXVIII. 
Find  the  L.  C.  M.  of 

1.  a362anda268.  7.  9  aSftSa^y  and  8  a^. 

2.  a6c2  and  a'^b(^.  8.  |  a^b^c^  and  2  ¥<fi. 

3.  9  ab^  and  6  a^b.  9.  42  axy^z^  and  77  6^- 

4.  4  y^y  and  10  xy^.  10.  ab^,  a%c,  and  abd^. 

5.  24a363a4  and  ma%^7^,  11.  Zx'^yz^,  \bxy^z%  and  \(ix^yH\ 

6.  a%H^  and  3  b^x.  12.  aft^c^x*,  a^bd^T?,  and  a^'^cx^. 

141.  When  the  factors  of  two  or  more  multinomial 
expressions  are  known,  their  L.  C.  M.  can  be  at  once 
written  down. 

The  L.  C.  M.  will  be  the  product  obtained  by  taking 
every  factor  which  occurs  in  the  different  exjwessions  to  the 
highest  power  which  it  has  in  any  one  of  them. 

Consider,  for  example,  the  expressions  (x  —  a)(x  —  by  (x  —  c)^ 
and  (x  —  ay  (x  —  6)  (x  —  c). 

It  is  clear  that  any  common  multiple  must  contain  (x  —  a)*  as  a 
factor  ;  it  must  also  contain  (x  —  by  and  (x  —  cy  as  factors.  Any 
common  multiple  must  therefore  contain  (x  —  a)*  (x  —  by  (x  —  cy 
as  a  factor;  and  hence  the  common  multiple  of  lowest  dimensions 
must  be  (x  —  a)*  (x  —  by  (x  —  cy. 


170  LOWEST   COMMON   MULTIPLES. 

Ex.  1.  Find  the  L.  C.  M.  of  a^V^  -  a%^  and  a^h^  +  a%^. 

Since  a^h"^  -  a^h^  =  a^b'^  («-&)(«+&), 

and  a*63  +  ^3^4  ^  ^353  (a  +  6) , 

we  see  that  the  L.  C.  M.  is  a^b^  (a  -&)(«  +  &). 

Ex.  2.  Find  the  L.C.M.  of  a^+Sa%-\-2ah'^2ind  a^+^a^b-{-Sa^b\ 
a3  +  3a26  +  2a62  =  «(«  +  &)(«  +  2  6), 
and  aj4  _^  4(^35  4.  3  ^2^,2  ^  q^2((^  4.  5)^^  +  3  5)  . 

hence  the  L.  C.  M.  is  a'^(a  +  6)  (a  +  2  6)  (a  +  3  &) . 

We  leave  the  L.  C.  M.  in  the  above  form,  as  it  is  generally  most 
convenient  to  have  it  expressed  in  factors. 

Ex.  3.  Find   the   L.  C.  M.   of    ^2  +  2a:  +  1,    ^2  -  2  x  -  3,   and 

x2  +  4x4-3. 

a;2  +  2a:+l=(a:  +  l)2, 

a;2_2x-3=(x+l)(x-3), 

and  x^  +  ^x  +  S=(x+l){x  +  S); 

hence  the  L.  C.  M.  is  (x  +  l)%x  -  3)(x  +  3). 

EXAMPLES    XXXIX.  _ 

Find  the  L.  C.  M.  of 

1.  {a  —  x)(a  —  2x)  and  (a  —  2ic)(a  —  3x). 

2.  ax^(a  —  x)(a  —  2x)  and  a%(a  —  2  x)  (a  —  3  x). 

3.  a2  -  62  and  (a  +  b)^. 

4.  6  «6(a  +  6)2  and  4  a2(Qj2  _  52). 

5.  x2  +  3x  +  2  and  x2  +  5x  +  4. 

6.  x2  +  3  xy  +  2  ?/2  and  x2  +  5  xy  +  4  y^. 

7.  x2  -  4  X  +  3  and  x2  -  5  x  +  6. 

8.  3  x2  -  4  xy2  +  ?/4  and  6  x2  -  5  x?/2  +  y*. 

9.  (a  +  6)2,  (a  -  6)2,  and  a^  -  b^. 

10.  (x  +  2  ?/)2,  (x  -  2  ?/)2,  and  x'^-iy^. 

11.  x2  +  7x  +  12,  x2  +  6x  +  8,  and  x2  +  5x  +  6. 

12.  x2  -  7 xy  +  12 2/2,  x2  -  6xy  +  8 y2,  and  x2  -  5xy  +  62^2. 


LOWEST  COMMON   MULTIPLES.  171 

142.  When  the  factors  of  the  expressions  whose 
L.  C.  M.  is  required  cannot  be  seen  by  inspection,  we 
must  use  the  rule  for  finding  the  H.  C.  F.  given  in  the 
preceding  chapter. 

Thus,  to  find  the  L.  C.  M.  of  a^  +  ^2  -  2  and  x^-\-2x'^-3. 
The  H, C.F.  of  the  given  expressions  is  x—  1,  and  we  find  by 
division  that 

a;3  +  x^  — 2  =(x  -  l)(a;2  +  2x  +  2), 

and  x^-h2x^-S=(x-  l)(a;2  +  3a:  +  3). 

Then,  since  x'^  -\-2x-\-2  and  x2  +  3  x  +  3  have  no  common  factors, 
the  L.  C.  M.  required  is 

'{X  -  l)(x2  -f  2a;  +  2)(x2  +  3x  +  3). 

143.  Let  A  and  B  stand'  for  any  two  algebraical 
expressions,  and  let  H  stand  for  their  H.  C.  F.,  and  L 
for  their  L.  C.  M. 

Let  a  and  h  be  the  quotients  when  A  and  B  respec- 
tively are  divided  by  /f;  so  that 

A  =  H  y^a, 
and  B=H  xb. 

Since  H  is  the  highest  common  factor  of  A  and  B,  a  and 
b  can  have  no  common  factors.  Hence  the  L.  C.  M.  of  A 
and  B  must  he  Hx  a  xb.     Thus 

L  =  H-a'b (i.). 

It  follows  from  (i.)  that 

L=.Hax^  =  Ax%     ....    (ii.), 

and  also  that 

LxH=HaxHb  =  AxB  .     .     .    (iii.). 


172  LOWEST   COMMON   MULTIPLES. 

From  (ii.)  we  see  that  the  L.  C.  M.  of  any  two  algebrai- 
cal expressions  is  found  by  dividing  one  of  the  expressions 
by  their  H.  C.  F.,  and  multiplying  the  quotient  by  the  other 
expression. 

From  (iii.)  we  see  that  the  product  of  any  two  expres- 
sions is  equal  to  the  product  of  their  H.  C.  F.  and  L.  C.  M.    - 

144.  To  find  the  L.  C.  M.  of  more  than  two  expres- 
sions, whose  factors  cannot  be  determined  by  inspection, 
we  first  find  the  L.  C.  M.  of  two  of  the  given  expressions, 
and  find  the  L.  C.  M.  of  this  result  and  of  the  third 
expression,  and  so  on. 

EXAMPLES  XL. 
Find  the  L.  C.  M.  of 

1.  3a26c,  27a^b^c:^,  and  6a6%.  3.    ia^  6a%  and  Sab^. 

2.  a^b%  b%  and  aV.  4.   2a2,  6ab%  and  4ta^b^. 

5.  x^(x  -  2/)2,  y^{x  +  ?/)2,  and  xy(x^  -  y^). 

6.  «»  +  a%  a(a-b),  and  a'^  -  b^. 

7.  xy'^  —  y^,  x^y^  +  xy'^,  and  x'^y  —  y. 

8.  2axy(x  -  y),  Sax:\x^  -  y2),  and  iy^^x  +  yy. 

9.  6(x2-9),  9(x  +  S),  15(aj-4),  and  10(a;2  -  x  -  12). 

10.  4  a  +  4  6,  6  a2  _  24  b'^,  and  a'^-Sab-\-2  62. 

11.  4  a63  _i_  4 1)3^^  0^-252  _  52(^2^  and  8  a^bd  -  8  abcP. 

12.  ^2  -  lOic  +  24,  x2  _  8 jc  +  12,  and  x'^-6x  +  8. 

13.  x2  -  9ic  -  10,  a;2  -  7a;  -  30,  and  x^  +  4a;  +  3. 

14.  2ic2-8,  3ic2-9x  +  6,  and  6x^  +  18x+  12. 

15.  x2-3ic  +  2,  2x2 -5C- 6,  and  3x2-2x-l. 

16.  6  x2  +  X  -  2,  21  x2  +  17x  +  2,  and  14x2  -  5x  -  1. 

17.  3x2  -  10 x?/  +  3 y^,  3 x2  -ixy  +  y^,  and  x2  -  4 xy  +  3 y^. 

18.  x8  +  2x2-3x  and  2x^  +  5a;2-3x. 


LOWEST  COMMON  MTJLTIPLES.  173 

19.  xV  _  9y2^  r^2y  _y^_Qy^  Siud  d^  +  x^  -  6  X. 

20.  x^-a^,  3i^-{-  a^,  and  a^  +  a'^x^  +  a*. 

21.  x2  -  1,  ic3  ^_  a;2  _f.  a;  ^  1^  and  a^  -  x2  +  a;  -  1. 

22.  9a^  -  x  -  2  and  Sic^  -  10x2  -  7x  -  4. 

23.  x8  -  ax2  -  a^x  +  a^  and  x^  +  ax2  -  a^x  -  a^, 

24.  x3  +  x2-4x-4  and  x34-6x2  + llx  +  6. 
26.  X*  -  x3  +  8 X  -  8  and  x3  +  4x2  -  8 X  +  24. 

26.  a^-{-6a^  +  na-hQ  and  a^  +  lOa^  +  29a  +  20. 


174  MISCELLANEOUS   THEOREMS. 


CHAPTER  XIII. 

Miscellaneous  Theorems  and  Examples. 

145.  Mathematical  Induction.  A- method  of  proof,  ordi- 
narily called  mathematical  inductioyi,  is  frequently  em- 
ployed in  the  demonstration  of  mathematical  propositions. 
The  method  is  best  explained  through  its  application  to 
simple  examples. 

Ex.  1.     The  following  equations  are  evidently  true : 
1  +  3  =   4  =  22,  ^ 
1  +  3  +  5  =    9  =  32,  ' 
1  +  3  +  5  +  7^=16  =  42, 

1  +  3  +  5  +  7  +  9  =  25  =  52;  5 

and  they  at  once  suggest  that  perhaps  the  sum  of  the  first  n  odd 
numbers  is  equal  to  n2.     Let  us  assume  provisionally  that  it  is. 
This  hypothesis  algebraically  expressed  is 

1  +  3  +  5  +  ...  +  2w-l  =  w2, 

in  which  n  is  an  integer  and  equal  to  the  number  of  terms  in  the 
sum. 

To  both  sides  of  this  equation  add  2  w  +  1.  The  result  of  the 
addition  is 

1  +  3  +  5  +  ...  +2w-l  +  2w  +  l  =  n2 +  2/1  +  1 

=  {n  +  1)2, 

and  shows  that  if  the  sum  of  the  first  n  odd  numbers  is  n2,  then 
the  sum  of  the  first  w  +  1  odd  numbers  is  (7i  +  1)2. 

But  the  formula  above  written  is  obviously  true  when  n  =  1  and 
when  n  =  2  ;  hence  it  is  true  when  n  =  3.    And  being  true  when 


MISCELLANEOUS  THEOREMS.  175 

n  =  3,  it  is  true  when  w  =  4,  and  so  on  indefinitely.  It  is  there- 
fore true  when  n  is  any  integer  whatever. 

Thus  the  proposition  that  the  sum  of  the  first  n  odd  integers  is 
equal  to  ri^  is  proved. 

Ex.  2.  Let  it  be  proposed  to  find  the  sum  of  the  first  n  natural 
numbers.  We  begin  by  writing  down  the  following  equations, 
which  are  evidently  true  : 

1  +  2  =  3   =  J.  2  (2+1), 

1  +  2  +  3  =  6    =^.3(3  +  1), 

1  +  2  +  3  +  4  =  10  =  ^.4(4  +  1), 

1  +  2  +  3  +  4  +  5  =  15  =  ^.5  (5+1). 

They  at  once  suggest  that  perhaps  the  sum  of  the  first  n  integers 
is  ^  n  (n  +  1).    We  accordingly  assume,  provisionally,  that 

l+2  +  3  +  4  +  ...  +  n  =  ^n(w  +  l). 

Adding  n+ 1  to  both  sides  of  this  hypothetical  equation,  we  obtain 

1  +  2  +  3  +  ... +  n  +  n  +  l  =  |w2  +  ^n  +  «  +  l 
=  i(,i  +  l)(n  +  2), 

a  result  which  shows  that  if  \n{n  -\-  V)  is  the  sum  of  the  first 
n  integers,  then  ^(w  +  l)(7i  +  2)  is  the  sum  of  the  first  n  +  1 
integers. 

But  the  formula  is  obviously  true  when  n  =  1 ;  hence  it  is  true 
when  11  =  2.  And  being  true  when  n  =  2,  it  is  true  when  n=  3, 
and  so  on  indefinitely.  It  is  therefore  true,  whatever  integer  n 
may  represent. 

These  examples  may  suffice  to  explain  a  method  of 
proof  that  will  be  made  use  of,  as  occasion  requires,  in 
subsequent  pages  of  this  book.  If  its  logical  rigor  be 
not  at  first  entirely  evident  to  the  student,  a  careful 
study  of  its  applications  in  a  few  examples  that  require 
its  use  should  remove  the  doubt. 

Ex.  3.    Prove  that  the  sum  of  the  first  n  even  integers  is  n(n  + 1). 


176  MISCELLANEOUS   THEOREMS. 

Ex.  4.  Prove  that  the  sum  of  the  squares  of  the  first  n  integers 
is^n(n  +  l)(2w+l). 

Ex.  6.     Prove  that 

I.2  +  2.3  +  3.4+...  +  w(w  +  l)=|n(w  +  l)(7i  +  2). 

146.  The  term  induction  does  not  correctly  describe  the 
above  method  of  proof.  In  the  natural  sciences  the  true 
method  of  induction  is  used  for  the  purpose  of  inferring 
the  universality  of  natural  laws  from  particular  mani- 
festations of  their  truth  in  natural  phenomena.  It  does 
not  seek  to  attain  absolute  certainty,  but  achieves  its 
object  as  soon  as  it  establishes  a  high  degree  of  proba- 
bility. 

The  so-called  method  of  mathematical  induction,  how- 
ever, is  as  logically  rigorous  as  any  other  process  in  mathe- 
matics. It  is  more  correctly  described  as  reasoning  by 
progressive  transformations,  in  which  the  uniqueness  of 
an  algebraic  form  persists,  —  monomorphic  transforma- 
tions, or  monomorphosis. 

147.  Theorem.  The  expression  ic"  —  a**  is  divisible  by 
X  —  a  for  all  positive  integral  values  of  n. 

It  is  known  that  x  —  a,x^  —  a^,  and  a^  —  a^  are  all  divisi- 
ble by  a;  —  a. 

We  have     x''  —  a'*  =  x'^  —  ax'^~'^  -f  ax^~^  —  a" 

=  x""-^  (x  —  a)-i-a  (x""-^  —  a'*-^). 
Now,  if  x  —  a  divides  cc*""^  —  a**"^,  it  will  also  divide 
a;n-i  (^x  —  a)-\-a  {x""-^  —  a""-^), 

that  is,  it  will  divide  x^  —  a\     (The  if  of  this  sentence 
is  important.) 


MISCELLANEOUS   THEOREMS.  177 

Hence,  if  x  —  a  divides  a;"~^  —  a"~^,  it  will  also  divide 
ic"  —  a". 

But  we  know  that  x  —  a  divides  oi?  —  a^;  it  will  there- 
fore also  divide  a;*  —  a*.  And,  since  x  —  a  divides  x^  —  a^, 
it  will  also  divide  ar^  —  a\     And  so  on  indefinitely. 

Hence  a;**  —  a"  is  divisible  by  x  —  a,  when  n  is  any  posi- 
tive integer  [Art.  145]. 

Since  x"  +  a"  =  af*  —  a"  +  2  a",  it  follows  that  when 
x^-\-  a*"  is  divided  by  a;  —  a  the  remainder  is  2a''y  so  that 
a;"  -f-  a"  is  never  divisible  by  a;  —  a. 

If  we  change  a  into  —a,  x—a  becomes  a;—  (—a)  =x+a; 
also  a;**  —  a"  becomes  af*  —  (  —  a)" ;  and  x"  —  (  —  a)"  is 
a;«  -f  a"  or  «"  —  a**  according  as  w  is  odd  or  even. 

Hence,  when  n  is  odd, 

x^  +  a**  is  divisible  by  a;  +  a, 
and  when  n  is  even, 

a;«  —  a"  is  divisible  by  a;  +  a. 
Thus,  when  n  is  a  positive  integer, 

x  —  a  divides  a;"  —  a"  always, 

x  —  a    .  .  .  .  af  -f  a"  never, 

x-\-a    ....  af  —  a**  when  n  is  even, 
and  x-{-a    ....  a;"  +  a"  ...  .  odd. 

We  have  in  the  above  shown  that  the  four  cases  are 
all  included  in  the  first :  we  leave  it  as  an  exercise  for 
the  student  to  prove  each  case  separately. 

The  above  results  may  be  written  so  as  to  show  the 
quotients :  thus 


178  MISCELLANEOUS   THEOREMS. 


—  -  x''-'^  +  x^'-^a  +  x^'-^a'  H h  a"-\ 


X—  a 

x""-^  —  x^~^  a  +  x^'-^a^ ±  a~-^ 


x""  ±  a        ._i       „_2 


x-}-  a 

the  upper  or  lower  signs  being  taken  on  each,  side  of  the 
second  formula  according  as  n  is  odd  or  even. 

EXAMPLES  XLI. 

1.  Find  the  factors  of  x^  —  x^  —  x  -{■  1. 

2.  Find  the  factors  of  c(fi  -  x^  -  x"^  +  1. 

3.  Find  the  factors  of  I  —  x  —  x^  +  x^. 

4.  Find  the  factors  of  1  —  x^  —  x^  +  x^^. 

5.  Find  the  factors  of  x^  -  y"^  -  2ax  -  2by -{■  a^  -  h\ 

6.  Find  the  factors  oi  x^  -  y'^  -Zx- y  ^2. 

7.  Find  the  factors  of  a-.2  -  2  xy  -  8  ?/2  _  2  x  +  20  ?/  -  8. 

8.  Find  the  factors  of  a^  -  62  +  c2  _  /2  _  2  ac  -  2  hf. 

9.  Write  down  the  quotient  in  each  of  the  following  divisions : 
(i.)  (x5  -  y^)  -  (X  -  y),  (ii.)  (x^  +  f)  -  (a;  +  y), 

(iii.)  {x^  -y'^^-^ix^  y). 

10.  Show  that,  if  n  is  any  positive  integer,  7"  —  1  is  divisible  by 
6  ;  show  also  that  SS^^+i  +  1  is  divisible  by  36.  . 

11.  Show  without  actual  division  that 

(3x2  -  2  a;  +  1)3 -(2a;2  +  3x  -  5)8 
is  divisible  by  x2  —  5  x  +  6. 

12.  Write  down  the  result  of  dividing  (2  a  +  3  6)8  +  (3  a  +2  6)8 
by  5  a  +  5  6. 

13.  Write  down  the  result  of  dividing 

(2  a  +  4  6  -  4  c)3  +  (a  -  6  +  7  c)8  by  a  +  6  +  c. 

14.  Show  that  (1  —  x)  2  is  a  factor  of  1  —  x  —  x°  +  x^. 


MISCELLANEOUS  THEOREMS.  179 

15.  Show  that  (1  —  xy  is  a  factor  of  1  —  x  —  x"  +  x'*+\  n  being 
any  positive  integer. 

16.  Show  that  (x  -  1)2  is  a  factor  of  wx"+i  -  (n  +  l)x"  +  1,  and 
also  of  x''  —  wx  +  w  —  1,  w  being  any  positive  integer. 

148.  Factor  Theorem.  If  any  rational  and  integral  ex- 
pression which  contains  x  vanish  when  f  is  put  for  x,  then 
will  x—fhe  a  factor  of  the  expression. 

Let  the  expression,  arranged  according  to  powers  of 
Xj  be 

ax''-\-  hx""-^  +  caf"-^ -\ 

Then,  by  supposition, 

Hence  aa;"  +  dx""^  +  cx""'^  -\ 

=  aaf»  +  &af*"^  +  ca;'*-^  h (a/"  +  ft/''-^  +  c/— ^  +  ..•) 

=  a  (x»  -/")  +  h  (a;"-i  —f~^)  +c  (af'-^  —f-^)  +  ... 
But,  by  the  last  article,  a;"-/**,  a;'-^-/^-!,  a;"-2_yn-2^ 
etc.,  are  all  divisible  by  a;  —  / 

Hence  also  ax""  +  hx""'^  +  cx""^  +  •••  is  divisible  by  x—f. 

149.  Eemainder  Theorem.  If  any  expression  which  con- 
tains X  he  divided  by  x—f  the  remainder  is  equal  to  the 
result  obtained  by  putting  f  in  the  place  of  x  in  the 
expression. 

Divide  the  expression  ax'*  4-  6af~^  +  cx"~^  -\ by  a;  — /, 

continuing  the  process  until  the  remainder,  if  there  be 
any  remainder,  does  not  contain  x\  and  let  Q  be  the 
quotient,  and  R  the  remainder. 

Then,  by  the  nature  of  division, 

asf"  +  6af-i  +  csf'-^  +  ...  =  Q(a;  -/)  +  R, 
and  the  two  members  of  this  equation  are  identical. 


180  MISCELLANEOUS   THEOREMS. 

Now  since  R  does  not  contain  x,  no  change  will  be 
made  in  R  by  changing  the  value  of  x\  put,  then,  x=f, 
and  we  have  the  required  result, 

ar  +  ¥"-'  +  cp-'  +  -  =  Q{f-f)  +  R  =  R. 
This  is  known  as  the  remainder  theorem.  It  includes 
the  factor  theorem  of  Art.  148  as  a  particular  case,  or 
corollary;  for,  a/'*  +  ^/^^-^  +  c/"-^  +  . . .  =  0  when  R=0, 
and  if  i?  =  0,  the  expression  ax""  -\-  hx^~^  +  cx""'^  +  •••  is 
exactly  divisible  by  x—f. 

Ex.  1.  Find  the  remainder  when  x^  —  4a;2  +  2  a;  +  1  is  divided 
by  a;  -  3. 

The  remainder  is  33  -  4  •  32  +  2  •  3  +  1  =  -  2. 

Ex.  2.    Show  that  x  -  2  is  a  factor  of  «*  -  3  a;2  +  2  a;  -  8. 
The  condition  is  2*  —  3- 22  +  2- 2-8  =  0;  and  this  condition 
is  satisfied. 

Ex.  3.  Determine  whether  aj^  —  3  aj  +  2  and  x^  —  13  a:  +  12  have 
a  common  factor. 

The  factors  of  a;^  —  3  a;  +  2  are  obviously  a;  —  1  and  a:  —  2  ;  and 
of  these  a;  —  1  does,  and  x  —  2  does  not,  satisfy  the  above  condition 
of  being  a  factor  of  a;^  —  13  x  +  12. 

Ex.  4.  Show  that  any  rational  and  integral  expression  in  x  is 
divisible  by  x  —  1  if  the  sum  of  the  coefficients  of  the  different 
powers  of  x  is  zero. 

For  if  the  sum  of  the  coefficients  is  zero,  the  expression  will 
vanish  when  we  put  x  =  1,  and  therefore  x  —  1  is  a  factor. 

150.  Symmetry.  An  expression  is  said  to  be  symmetri- 
cal with  respect  to  any  two  letters  when  it  is  unaltered 
by  an  interchange  of  the  two  letters. 

Thus  a  +  6  and  a^  +  2>^  are  symmetrical  with  respect 
to  a  and  6,  for  neither  expression  would  be  altered  by 
changing  a  into  h  and  h  into  a. 


MISCELLANEOUS   THEOREMS.  181 

Also  a  +  b  -{-c  and  a^  -{-b^  -\-c^  —  Sabc  are  symmetrical 
with,  respect  to  any  two  of  the  three  letters  a,  b,  c. 

The  only  expression  of  the  first  degree  which  is 
symmetrical  with  respect  to  the  three  letters  a,  b,  c 
is  pa-\-pb+pc,  where  p  is  some  numerical  coefficient. 

These  examples  suggest  the  following  definition : 

Def.  An  expression  is  (completely)  symmetrical  with 
respect  to  a  specified  set  of  letters  when  it  is  symmetri- 
cal with  respect  to  every  pair  of  them. 

Ex.  The  following  expressions  are  symmetrical  with  respect  to 
the  letters  contained  in  them  : 

a2  +  62  +  c2  -  6c  -  ca  -  ab, 
(6-c)4+(c-a)*  +  (a-6)*, 
bed  +  cda  +  dab  +  abc. 

151.  A  partial  symmetry  may  exist  in  expressions 
containing  more  than  two  letters.  Thus,  although 
ab^  -i-bc^  -\-  ca^  is  not  completely  symmetrical,  but  in 
fact  changes  its  form  when  any  two  of  its  letters  are 
interchanged,  yet  it  remains  unaltered  when  a  is  changed 
to  b,  b  to  c,  and  c  to  a.  Such  a  series  of  changes  is 
called  a  cyclic  displacement,  and  the  symmetry,  repre- 
sented in  such  an  expression  as  ab^  -\-bc^-\-  ca^,  is  called 
cycle-symmetry. 

Def.  An  expression  is  cyclo-symmetrical  with  respect 
to  the  letters  a,  b,  c,  cZ,  •  •  •  Z  if  it  remains  unaltered  when 
a  is  changed  to  6,  6  to  c,  c  to  d,  etc.,  and  I  to  a. 

Ex.  The  following  expressions  are  cyclo-symmetrical  with 
respect  to  the  letters  contained  in  them: 

a\b  -  c)  +  62(c  -  a)  +  c2(a  -  6), 

(6-c)3  +  (c-a)8+(a-6)8, 

(6  -  c  -  l)(c  -  a  -  l)(a  -  6  -  1). 


182  MiSCELLAi^EOtJS   THEOREMS. 

152.  The  arrangement  of  the  terms  in  a  symmetrical 
expression  is  of  some  importance.  Consider,  for  example, 
the  arrangement  of  the  expression  bc-\-  ca-\-  ah.  The 
term  which  does  not  contain  the  letter  a  is  put  first,  and 
the  other  terms  can  be  obtained  in  succession  by  chang- 
ing a  into  h,  b  into  c,  and  c  into  a.  In  the  expression 
a\b  —  c)  -f  6^(c  —  a)  -f  c^^a  —  b)  the  same  arrangement  is 
observed ;  for  by  a  cyclic  displacement  of  the  letters  in 
a^{b  —  c)  we  obtain  b'\c  —  a),  and  another  cyclic  change 
will  give  c^{a  —  b). 

By  reason  of  this  law  of  the  derivation  of  terms 
from  one  typical  term,  symmetrical  expressions  are  fre- 
quently not  written  out  in  full,  but  indicated  by  placing 
the  Greek  letter  S  before  the  typical  term.  Thus,  when 
two  letters  are. involved, 

^a'b  =  a'b  +  b% 

or,  if  three  letters  are  involved, 

^a'b  =  a'b  -\-  ah  -f  b'^c  +  b'^a  -f  c^a  +  (?b. 

Similarly,  when  three  letters  are  involved, 

2a.2  +  %bc  =  a^  +  52  _|_  c^  _l_  5c  -f  ca  +  ab. 

153.  We  now  proceed  to  consider  some  examples 
which  will  illustrate  the  theorems  proved  in  the  pre- 
ceding articles. 

Ex.  1.  Show  that,  if  w  be  a  positive  integer,  x"  —  2/**  is  divisible 
by  X  -  y. 

If  X  =  2/,  then 

ajn  _  yn  _  gjn  _  /jjn  —  Q^ 

Hence  the  proposition  is  an  immediate  consequence  of  the  factor 
theorem,  which  is  itself  a  corollary  of  the  remainder  theorem. 


MISCELLANEOUS   THEOREMS.  183 

•  Ex.  2.    Show  that    (&  -  cy  +  (c  -  ay  +  (a  -  by 
is  divisible  by  (&  —  c)(c  —  a)(a  —  6). 

If  we  put  6  =  c  in  the  expression 

(6  -  cy  +  (c  -  ay  +  (a  -  by, 
the  result  is  (c  —  a)^  +  (a  —  c)^ 

which  is  zero. 

Hence  (6  —  c)  is  a  factor,  and  we  can  prove  in  a  similar  manner 
that  c  —  a  and  a  —  b  are  factors. 

Ex.  3.     Show  that 
(a  -{■  b  -h  cy  -(b  +  c  -  ay  -(c  +  a  -by-ia  +  b  -  cy  =  24 abc. 

If  we  put  a  =  0  in  the  expression 

(a  +  &  +  c)8  - (&  +  c  -  ay  -(c  +  a-by-(a  +  b-  cy  (i.), 
it  is  easy  to  see  that  the  result  will  vanish  for  all  values  of  b  and  c. 

Hence  a  is  a  factor  of  (i.). 

So  also  b  and  c  are  factors  of  (i.). 

Now  (i.)  is  an  expression  of  the  third  degree ;  it  can  therefore 
only  have  three  factors.  Hence  it  is  either  equal  to  abc,  or  is 
equal  to  abc  multiplied  by  some  number. 

Thus 

(a  +  b  +  cy-(b-^c  -  ay  -(c-^  a  -  by-(a  +  b  -  cy  =Labc(u.'), 
where  L  is  some  number  which  is  always  the  same  whatever  a,  6, 
and  c  may  be. 

We  can  find  the  value  of  L  by  giving  particular  values  to  a,  6, 
and  c. 

Thus,  let  a  =  6  =  c  =  1.     Then  (ii.)  becomes 
38  _  18  _  18  _  18  ^  X. 

.-.  X  =  24. 
Ex.  4.     Find  the  factors  of 

a8  (6  -  c)  +  b^  (c-a)+  c«  (a  -  6). 
If  we  put  6  =  c  in  the  expression 

a3  (6  -  c)  +  68  {c-a)+  c*  (a  -  6), 
the  result  is  <fi  (c  —  a)-{-  c^  (a  —  c), 

whinh  is  clearly  zero. 


184  MISCELLANEOUS   THEOREMS. 

Hence  6  —  c  is  a  factor  of  the  given  expression ;  and  we  can 
prove  in  a  similar  manner  that  c  —  a  and  a  —  h  are  also  factors. 

Now  the  given  expression  is  of  the  fourth  degree ;  hence, 
besides  the  three  factors  we  have  found,  there  must  be  one  other 
factor  of  the  first  degree,  and  as  this  factor  must  be  symmetrical 
in  a,  6,  c,  it  must  be  a  +  ft  -f  c. 

Hence  the  given  expression  must  be  equal  to 

i  (&  -  c)  (c  -  a)  (a  -  6)  (a  +  &  +  c), 
where  i  is  a  number. 

We  can  find  L  by  giving  particular  values  to  a,  6,  and  c  ;  or,  by 
comparing  the  coefficients  of  a^,  we  have  at  once 

h  -c  =  -L{h-c). 
Hence  L  =—\. 

Thus    S  53  (c  _  a)  =  _  (6  -  c) (c  -  a)(a  -  &) (a  +  6  +  c). 

154.    The  following  is  an  important  identity : 
d^^W^c^-^ahc  =  (a  +  &  +  c)  (a^  -^-y  -\-c^  -  he  -  ca-  ah). 

It  should  be  noticed  that 
a?j^W  J{.  c"  -he  -  ca-ah=\\{h-cy+  {c-ay+  {a-hy\. 

Since  a-\-h-\-c  is  a  factor  of  a^  +  6^  +  c^  —  3  ahc,  it 
follows  that  a-^  4-  &'  +  c'  -  3a6c  =  0  if  a  +  6  +  c  =  0. 

Hence  o?  -\-W  +  (?=3  ahc  for  all  values  of  a,  h,  and  c, 
provided  only  that  a  4-  &  +  c  =  0. 

That  is,  the  sum  of  the  euhes  of  any  three  quantities  is 
equal  to  three  times  their  product,  provided  that  the  sum  of 
the  three  quantities  is  zero. 

For  example,  the  letters  involved  being  a,  6,  c, 
2  (&  -  c)3  =  8  (&  -  6) (c  -«)(«-&), 
2  (ft  +  c  -  2  a)3  =  3  (6  +  c  -  2  a)  (c  +  a  -  2  6)  (a  +  6  -  2  c)  ; 
also,  since  {x  -  a)(h  -  c)-{-{x  -  h) (c  -  a)  +  (x -  c) (a  —  6)  =  0, 
we  have 
2  (a;  -  a)3  (6  _  c)8  =  3  (x  -  a)(x  -  6)(ic  -  c)(h-c){G  -a){a  -  6). 


MISCELLANEOUS   THEOKEMS.  185 

EXAMPLES  XLII. 

Let  the  student  employ  the  2  notation  for  the  purpose  of  abbre- 
viating such  of  the  following  expressions  as  admit  of  its  use. 

1.  Show  that      a2  (&  _  c)  +  &2  (^c-  a)+  c^  (a  -  ft) 

=  6c  (6  -  c)  +  ca  (c  —  a)  +  ab  (a  —6)  =  -  (6  -  c)  (c  -  a)  (a  - 6). 

2.  Show  that      a^  (6  -  c)+ &*  (c  -  a)+ c*  (a  -  6) 

=  -(6  -  c)(c  -  a)(a  -  &)(a2  +  52  _|_  c2  +  &c  +  ca  +  a&). 

8.   Show  that      (6  -  c)^  +  (c  -  ay  +  (a  -  6)5 
=  5  (6  -  c)(c  -  a)(a  -  6)(a2  +  ^a  _^.  c2  _  ^c  -  ca  -  a6). 

4.    Show  that 
(a  +  6  +  c)*  -(6  +  cy  -  (c  +  ay  -  (a  +  6)*  +  a*  +  6*  +  c* 
=  12  a6c  (a  +  6  +  c). 

6.   Show  that      a  (6  -  c)3  +  6  (c  -  ay  -{- c  (a  -  by 
=  {b- c){c- a)(a-b){a  +  b  +  c). 

6.  Show  that      62c2  (b-c)+  c^a'^  {c-a)-{-  a^b'^  (a  -  6) 
=  —  (6  —  c)  (c  —  a)  (a  —  6)  (6c  +  ca+  ab) . 

7.  Show  that      a»  (62  -  c2)  +  68  (c2  -  a2)  +  c^  (a2  -  62) 
=  —  (6  —  c)  (c  —  a)  (a  —  6)  (6c  +  ca  +  ab). 

8.  Show  that      a*  (62  -  c2)  +  6*  (c2  -  a2)  +  c^  (a2  -  62) 
^-(6  -\-c){c  +  a)(a-\-  6)(6  -  c)(c  -  a)(a  -  6). 

9.  Find  the  factors  of  6c  (62  -  c2)  +  ca  (c2  -  a2)  +  a6  (a2  -  62). 

10.  Find  the  factors  of  a  (6  +  c  -  a)2  +  6(c+a-6)2+c(a  +  6-c)2 

+  (6  +  c  -  a)  (c  +  a  -  6)  (a  +  6  -  c). 

11.  Find  the  factors  of  a2  (6  +  c  -  a)  +  62  (c+a  -  6)  +  c2  (a+6-c) 

-  (6  +  c  -  a)  (c  +  a  -  6)  (a  +  6  -  c). 

12.  Find  the  factors  of  a  (6  +  c)  (62  +  c2  -  a2) 

+  6  (c  +  a)(c2  +  a2  -  62)+  c  (a  +  6) (a2  +  62  -  c2). 

18.   Find  the  factors  of 

(y  -  «)^  +  (2  -  x)^  -{-{x  -  y)^  -  k{y  -z){z-  x){x  -  y). 

14.   Find  the  factors  of 

a6(62  _  c2)3  +  56(c2  _  ^2)8  4.  c6(a2  _  62)8. 


186 


FRACTIONS. 


CHAPTER   XIV. 
Fractions. 

155.  To  obtain  the  arithmetical  fraction  ^,  we  must 
divide  the  unit  into  7  equal  parts  and  take  5  of  those 

parts.     So  also  to  obtain  the  fraction  -,  where  a  and  b 

h 

are  positive  integers,  we  must  divide  the  unit  into  b  equal 

parts  and  take  a  of  those  parts. 

156.  The  numerator  and  the  denominator  of  a  frac- 
tion, defined  as  in  Art.  155,  must  both  be  positive  inte- 
gers :  we  cannot,  for  example,  with  that  definition,  have 

3 

such  a  fraction  as  —^ ;  for  to  say  that  the  unit  is  to  be 

divided  into  —  f  equal  parts,  and  that  |  of  such  parts 
are  to  be  taken,  is  without  meaning. 

We  must  therefore  suppose  the  letters  in  -  to  be  re- 
ft 

stricted  to  positive  integral  values,  or  we  must  alter  the 

definition  of  - ;  and  as  we  cannot  restrict  the  values  of 
b 

the  letters,  we  must  entirely  dispense  with  the  fractional 

form,  or  make  some  modification  in  its  meaning. 

Now,  with  the  definition  of  Art.  155,  to  multiply  the 

fraction  -  by  6,  we  must  take  each  of  the  a  parts  b 


FRACTIONS.  187 

times ;  we  thus  get  ab  parts,  the  parts  being  such  that 
each  b  of  them  make  up  a  unit,  and  therefore  the  whole 
ab  parts  will  make  up  a  units.     Thus 

^Xb==a      ......     (i.). 

Dividing  each  side  by  b,  we  have 

g=«^6 («•)• 

Now  we  may  define  the  fraction  -  as  that  quantity  which 

when  multiplied  by  b  becomes  a;  for,  as  we  have  just 
seen,  this  new  definition  agrees  with  that  of  Art.  155, 
whenever  the  definition  of  Art.  155  has  meaning ;  and  by 
taking  this  new  definition  we  do  away  with  the  necessity 
of  ascribing  only  positive  integral  values  to  the  letters. 

We  may  similarly  dejirie  the  fraction  -  as  the  quotient 

ob'tained  by  dividing  a  by  b. 

Hence,  instead  of  the  definition  of  Art.  155,  which  is 
inapplicable  to  an  algebraical  fraction,  we  have  either  of 
the  following  equivalent  definitions. 

Def.  I.     The  algebraical  fraction  -,  where  a  and  b  are 

b 

supposed  to  have  any  values  whatever,  is   that  quantity 

which,  when  multiplied  by  b,  becomes  equal  to  a. 

Def.  II.  The  algebraical  fraction  -  is  the  quotient 
obtained  by  dividing  a  by  b. 

The  fractional  form  —  has  already  been  used  with  the 
b 


188  FRACTIONS. 

meaning  a-^b,  and  henceforth  the  notation  a/b  will  also 
be  frequently  employed  to  denote  a  fraction. 

157.  We  now  proceed  to  consider  the  properties  of 
algebraical  fractions ;  and  we  shall  find  that  algebraic:il 
fractions  are  added,  subtracted,  multiplied,  divided,  aiid 
simplified,  precisely  in  the  same  way  as  arithmetical 
fractions.  It  will  be  assumed  throughout  the  discussion 
that  the  quantities  involved  are  all  finite  and  different 
from  zero. 

158.  The  value  of  a  fractmi  is  not  altered  by  multiply- 
ing its  numerator  and  denominator  by  the  same  quantity. 

We  have  to  prove  that 

a_am 
b      bm' 
for  all  values  of  a,  &,  and  m. 

Let  a;  =  -.     Then  xxb  =  -xb. 
b  b 

But  -  X  6  =  a,  by  definition. 

b  ' 

.:  xb  =  a',  and  therefore  xbm  =  am. 

Divide  by  bm,  and  we  have  x  =  am  -j-  bm ; 

.  r   ,  •  a     am 

that  IS,  -  = 

b      bm 

Thus  the  value  of  a  fraction  is  not  altered  by  multiplying 
its  numerator  and  denominator  by  the  same  quantity. 

159.  Since,  by  the  last  article,  the  value  of  a  fraction 
is  not  altered  by  multiplying  both  the  numerator  and  the 
denominator  by  the  same  quantity,  it  follows  conversely 


FRACTIONS.  189 

that  the  value  of  a  fraction  is  not  altered  by  dividing 
both  the  numerator  and  the  denominator  by  the  same 
quantity. 

Hence  a  fraction  may  be  simplified  by  the  rejection  of 
any  factor  which  is  common  to  its  numerator  and  denom- 
inator. For  example,  the  fraction  a^y/a^y  takes  the  sim- 
pler form  aya^,  when  the  factor  y,  which  is  common  to 
its  numerator  and  denominator,  is  rejected. 

When  the  numerator  and  denominator  of  a  fraction 
have  no  common  factors,  the  fraction  is  said  to  be  in  its 
lowest  terms. 

160.  Eeduction  of  Tractions  to  their  Lowest  Terms.  To 
reduce  a  fraction  to  its  lowest  terms,  we  must  divide 
its  numerator  and  denominator  by  their  H.  C.  F. ;  for  we 
thus  obtain  an  equivalent  fraction  whose  numerator  and 
denominator  have  no  common  factors. 

Ex.  1.     Reduce  ^  to  its  lowest  terms. 

6  a'^xy 

The  H.  C.  F.  of  the  numerator  and  denominator  is  3  axy  ;  and 

Zax^y  _  3  ax^y  -f-  3  axy  _  _x_ 

6 a^xy     6 a'hcy  -h  3 axy  ~  2a 

Ex.  2.    Reduce  ^^  to  its  lowest  terms. 
a^b^ 

The  H.  C.  F.  of  the  numerator  and  denominator  is  a^ft* ;  and 

a^b*     a^b*  ^  a^b^     a^b 

Ex.  3.     Reduce  ^  «^&^^y^  to  its  lowest  terms. 
3  ab^x^y'^ 

The  H.  C.  F.  of  the  numerator  and  denominator  is  ab'^xy^  ;  and 

2  aWxy^  _  2  a^b'^xy^  ^  ab^xy^  _  2jcfiy^ 

3  ab^x^y^      3  abH^y^  -h  ab'^xy^      3  bx^ ' 


190  FRACTIONS. 

161.  Instead  of  reducing  a  fraction  to  its  lowest  terms 
by  dividing  the  numerator  and  denominator  by  their 
H.  C.  F.,  we  may  divide  by  any  common  factor,  and 
repeat  the  process  until  the  fraction  is  reduced  to  its 
lowest  terms. 

Thus  a^6^c3  ^  ^253^3  ^  giffi  ^  (jfl 

a^b*c^       6*0*        6c*      be 
The  above  process  may  be  written  down  more  compactly  as 
follows :  a^ 

a^d^c*     be 
be 

EXAMPLES  XLIII. 

Reduce  the  following  fractions  to  their  simplest  forms : 

-     o^  -     15  aWcPx^ 

ab"^'  '   26a'^¥c^x^' 

2    ^^  g    125  a6%3# 

xY  '    150a46W 

2  a^be^  g    3  a^x^yz-^ 

4  a^b^c  5  a¥xy^z 

^    6  a^b^c*  jQ   '5  a%^&xy^ 

4a765c4*  •     7b*c^xy^  ' 

6    ^!^.  11     14  ab'^c^xy'^z^ 

x'52/2^2'  •   21  a^b^cx^y^z 

g    12a;Vgio  j2     3  a^bc^x^y'^z 

16x^y^z^'  '  ^a^b^cx^y^z^' 

162.  When  the  numerator  and  denominator  of  a  frac- 
tion are  multinomial  expressions  whose  factors  can  be 
seen  by  inspection,  write  the  numerator  and  denominator 
as  the  product  of  factors  of  the  lowest  possible  dimen- 
sions; the  factors  which  are  common  to  the  numerator 


3. 


FRACTIONS.  191 

and   denominator   will   tlien   be   obvious,   and    can   be 

removed. 

,  ^    a^  —  ax 
Ex.  1.     Simplify  ^TT^' 

a^  —  ax  _      aja  —  x)      _     a 


Ex.  2.     Simplify 


a2  -  x2      {^^  x){a  +  x)     a  +  x 
x^  —  X- 


x*-l 

x*-!       (x»'-l)(x2  +  i3      a;2^1* 
a;2  _  7  X  +  10 


Ex.3.     Simplify   ^,_,^^g 

a;2  -  7 X  +  10  ^  (a;  -  5)(x  -  2;  ^  X  -  5 
x2_5x  +  6       (x-3)(x  — 2)      x-3' 

Ex.4.     Simplify    |5^. 

x''  —  ax  _       x(x  —  q) 
a2  _  x2     (o  -  x)(a  +  «)* 
Now  X  —  a  =  —  (a  —  x)  ;   hence,  dividing  the  numerator  and 
denominator  by  a  —  x,  we  have  the  equivalent  fraction, 


o  +  X 
If  we  divide  the  numerator  and  denominator  by  x  —  a,  we  have 

X 

-(a  +  x)* 

By  the  Law  of  Signs  in  Division    ~  ^    and are  both 

^  -a  +  x  -{a  +  x) 

equal  to ; — .  and  this  latter  is  the  form  in  which  the  result  is 

a  +  x' 

usually  left. 

Note.  — It  should  be  remarked  that  the  value  of  a  fraction  is  not 
altered  by  changing  the  signs  of  all  the  terms  in  the  numerator 
and  also  of  all  the  terms  in  the  denominator ;  for  this  is  equivalent 
to  multiplying  both  numerator  and  denominator  by  —  1. 


192  FRACTIONS. 

EXAMPLES  XLIV. 
Simplify 

^       2ab  g    4x- 16  ^„    6ax- 15  g^ 


18. 

3^    u_^^^^  jj_    j^iZlil.  19. 


21. 


a2  +  ab 

a^V 

10. 

(k2  -  x^y'^ 

a?--  ah 
a2  +  ah 

11. 

ic2-f  aa; 
a;2-a2 

12. 

(X  -  1)2 
X2-1 

13. 

X2  -  x2?/2 

(a:  +  xy)2 

14. 

x2+2x 

15 

x2-4 

x*  +  a:2 
x^-l 

16. 

x2  -  9  X  +  20 

x2  +  6x- 

55 

1-92/2  4-20?/* 

1+6^/2- 

55^4 

x2  -  8  x?/ 

+  72/2 

x2  —  3  x?/  ■ 

-282/2 

1  _  8  a2?,2  +  7  05454 

X2- 

-16 

2x3 

-4x* 

X2. 

-4x4 

X  — 

■2 

4- 

X2 

a  ~ 

-3 

9- 

a2 

a2x^ 

'-X* 

x* 

-a* 

X4- 

-^2 

a2- 

-ax2 

X5- 

-a2x3 

x* 

-a* 

^2x2 

2/2  -  x^y^ 

x-^  -  y  a;^ 

3  x2  -  12  ffx 

48a2_3a;2' 

a2  -  2  ax  4-  x2 
a2  _  x2 

rt4  +  2  «2/,2  4.  54 


-1 


X3-1 
X4-1 


23. 


24. 


X6-  1 

x2  -  5  X  +  6 
x2  -  7  X  +  12' 
1  -  5  a  +  6  a2 


30. 

27.     -^    -  -^^y  -^  >  y    .  3j 


x*2/4  -  a*  1  -  7  a  +  12  a2 

25    a;2-9x  +  20  gg     (g^  _  a;3) (a  +  x) 

(q3-68)(a2_(^5  4.52) 

{a^-h'^){a  +  h) 

(a^-h^)(a-b)  _ 
1  -  3  a262  _  28  a464  "     (^3  _  63)(^a4  _  ^4)* 

163.  When  the  factors  of  the  mimerator  and  denom- 
inator of  a  fraction  cannot  be  found  by  inspection,  their 
H.  C.  F.  can  be  fonnd  by  the  rule  given  in  Chapter  XI. ; 
and  the  fraction  will  be  reduced  to  its  simplest  form  by 
dividing  the  numerator  and  denominator  by  their  H.  C.  F. 


FRACTIONS.  193 

Thus  to  simply  ^-^^x  +  10^ 

The  H.  C.  F.  will  be  found  to  be  x^  -  5  x  +  2,  and 
a:8  _  23x  +  10  =(x2  -  5x  +  2)(x  +  5), 
5x3  -  23 x'^  +  4  =(x2  -  5x  +  2)(5x  +  2). 
Hence  the  given  fraction  is  equal  to 

x  +  5 
5x  +  2* 

EXAMPLES  XLV. 
Simplify 

J      x8-3x  +  2  g    2x8  +  3x2  +  4x-3 

2x8-3x2  +  1*  '         6x8  +  x2-l 

2        3x2-8x+5  ^  x8-3x-2 

x8-4x2  +  5x-2"  ■    x*  +  2x3  +  2x2  +  2x+ 1* 

g     x8  +  3x2-20  g  X*  -  x8  -  X  +  1 

x*  -  x2  -  12  '  '    X*  -  2x8  -  x2  -  2  X  +  l' 

^    x8  +  6x2+  llx  +  6  g    x*  +  2x8-3x2-7x-2 
x8  +  6x2  +  6x     '  *  2x*  +  x8-6x2-6x-l* 

g    2x8  +  ax2  +  4«2a;-7a8  ^^         x* -20x2  -  15x  +  4 
*   x8-7ax2  +  8a2x-2a8'        '   x*  +  9x3  +  19x2 -9x  -  20* 

164.  Eeduction  of  Fractions  to  a  Oommon  Denominator. 
Since  the  value  of  a  fraction  is  unaltered  by  multiplying 
its  numerator  and  denominator  by  the  same  quantity 
[Art.  158],  any  number  of  fractions  can  be  reduced  to 
equivalent  fractions,  all  of  which  have  the  same  denomi- 
nator. 

The  process  is  precisely  the  same  as  in  arithmetic,  and 
is  as  follows. 

First,  find  the  L.  C.  M.  of  all  the  denominators ;  then 
divide   the  L.  C.  M.  by  the  denominator  of  one  of  the 


194  FRACTIONS. 

fractions,  and  multiply  the  numerator  and  denominator 
of  that  fraction  by  the  quotient ;  and  deal  in  a  similar 
manner  with  all  the  other  fractions  :  we  thus  obtain  frac- 
tions equal  to  the  given  fractions,  all  of  which  have  the 
L.  C.  M.  for  denominator. 


For  example,  to  reduce 
X  y 


,  and 


a%{x  +  ay  ab^ix-a)'  ah  {x^  -  a^) 

to  a  common  denominator. 

The  L.  C.  M.  of  the  denominators  is  a^js  (-3,2  _  ^2-)_  Dividing  the 
L.  C.  M.  by  (jfih  (aj  -fa),  ah"^  (x  —  a),  and  ah  {od^  —  a"^),  we  have  the 
quotients  &  (x—  a),  a{x-\-  a),  and  ah  respectively.    Hence 

XX  h{x  —  a)  _    hx(x  —  a) 


and 


a'^h  (x  +  a)     a%  (x  +  a)xh(x-a)     a^h"-  (x^  -  a'^y 

y         _         y  X  a{x  +  a)         _    ay  {x  +  a) 
ah^  {X  -  a)     ah^  (^x  -  a)  x  a^x  +  a)     a%'^  (a;2  -  a^)' 

z  z  X  ah  ahz 


ah  (x2  -  a2)     ah  (x^  —  a^)  x  ah     a^h'^  {x?  -  a^) 


Note.  —  In  the  above  process  it  is  not  absolutely  necessary  to 
take  the  lowest  common  multiple  of  the  denominators ;  any  com- 
mon multiple  will  do  equally  well,  but  by  using  the  L.  C.  M.  there 
is  a  saving  of  labor. 

165.  Addition  of  Fractions.  The  sum  (or  difference)  of 
two  fractions  which  have  the  same  denominator  is  a 
fraction  whose  numerator  is  the  sum  (or  difference)  of 
their  numerators,  and  which  has  the  common  denomi- 
nator.    This  follows  from  Art.  80. 

When  two  fractions  have  not  the  same  denominator, 
they  must  first  be  reduced,  by  the  method  explained  in 
the  last  article,  to  equivalent  fractions  which  have  the 


FRACTIONS.  195 

same  denominator :  their  sum  or  difference  will  tlien  be 
found  by  taking  the  sum  or  difference  of  their  numera- 
tors, retaining  the  common  denominator. 

Thus  «  +  ^  =  ^+-^,  -* 

XX  X 

a_b_ a  —  b 

XX  X 

q,b_  a  X  y     b  x  x  _ay  ,bx_  ay  -f  bx 
X     y     XX  y     y  X  X     xy     xy         xy 

and  a  +  ^=:«l^  +  ^  =  «|ML&. 

y     y     y        y 

When  more  than  tWo  fractions  are  to  be  added,  or 
when  there  are  several  fractions,  some  of  which  are  to 
be  added  and  the  others  subtracted,  the  process  is  pre- 
cisely the  same.  The  fractions  are  first  reduced  to  a 
common  denominator,  and  then  the  numerators  of  the 
reduced  fractions  are  added  or  subtracted  as  may  be 
required. 

Thus  qj^b_c_^a  +  b-c^ 

^        ^        JO  tJu 

and  ^  +  ^  _  ^  =  a  X  yx     b  x  xz  _  c  x 


X     y     z     XX  yz      y  x  xz     z  x  xy 

_ayz  .  bxz  _  cxy  _  ayz  +  bxz  —  cxy 
xyz     xyz     xyz  xyz 

Note.  —  It  may  be  necessary  to  remind  the  student  that  when 
there  is  no  sign  between  a  fraction  and  a  letter  or  number,  tho 

sign  of  multiplication  is  understood.    Thus,  2-  means  2  x  -  and 

b  b 

not  2  -f  -  as  in  the  arithmetical  form  2i. 


196  FRACTIONS. 


Ex.  1.    Find  the  value  of  — h     ^ 


X  —  a     X  +  a 

The  L.  C.  M.  of  the  denominators  is  (x  —  a)(x  +  a);  and 
1,1_  x  +  a  X  —  a 


x  —  a     x-\-  a     {x  —  a){x  +  a)      {x  +  a){x  —  a) 
_a;  +  a-\-  x  —  a  _     2a; 

Ex.  2.    Find  the  value  of  — ^—  +  —^ — 
a  —  xx^  —  a^ 

Beginners  should  always  see  that  the  denominators  of  the  frac- 
tions which  are  to  be  added  are  all  arranged  according  to  descend- 
ing powers,  or  all  according  to  ascending  powers,  of  some  particular 
letter.     This  is  not  the  case  in  the  present  example ;  hut 
ax     _         ax        _   —  ax 

X'i  —  a^~-  (a-2  _  a;2)  ~  «2  _  x^" 

We  then  have 


a      \     ~  ^^  —  ^(^  +  ^^1  I     —  (i^ 

_  a{a  +  x)—  ax 
~        a-2  -  ic2 

Ex.  3.     Simplify  -J—  +  —1—  +  — ^  +  -  ^ 


V 


1-X       l-\-X       1  +  X2        1+X* 

It  is  sometimes  best  not  to  add  all  the  fractions  at  once ;  this 
is  particularly  the  case  when  the  denominators  are  not  all  of  the 
same  dimensions. 

1  1     ^(l  +  a;)  +  (l-a;)^      2      . 

1  -  X     1+  X        (1  -  X)  (1  +  x)        1  -  x2  ' 


then 


2(1 +x2)+2(l-x2)_      4 


1-X2        l-\-X^  (1-X2)(1  +  X2)  1-X*' 

and  finally 

4  4      ^  4(l  +  x*)+4(l-xO  ^      8 

1  _  a4  "•"  1  +  x*  (1  -  X*)  (1  4-  X*)  1  -  x8* 


FRACTIONS.  197 


Ex.  4.    Find  the  value  of  — — ^  +     ^ 


x-1     x+l      x-2     x+2 
Here  again  it  is  best  not  to  reduce  all  the  fractions  to  a  common 
denominator  at  once :   the  work  is  simplified  by  proceeding  as 
under. 

ce+l-(a;-l) 


x~l     x  +  1      (x-l)(x+l)      x^-l' 

1 1_  ^  x  +  2-(x-2)  ^     4 

x~2     x  +  2      (x-2)(x  +  2)      a;2-4' 
and  2  4      ^2(a;2-4)+4(x2-l)^     6a;2-12 

a;2_i     aj2-4  (x2-l)(x2-4)  a:* -6x2 +  4 

Ex.  6.    Find  the  value  of  ^ 


x2-Sx-\-6     a;2_7a.  +  i2' 

1 


x^-bx  +  Q      (a;-2)(x-3) 
and  1  1    * 


a;2_7x  +  12      (x-3)(x-4)' 

hence  the  L.  C.  M.  of  the  denominator  is  (x  —  2)(x  —  3)(«  —  4). 
Hence  we  have 

x-4  x-2 


(«-2)(ic-3)(ic-4)      (x-2)(x-3)(x-4) 

(X  -  4)  -  (x  -  2)  -2 

(x-2)(x-3)(x-4)      (x-2)(x-3)(x-4) 

2 

(x-2)(x-3)(x-4)* 

Ex.  6.   Find  the  value  of 


(o-6)(a-c)      (6-a)(6-c)      (c-a)(c-6) 

In  examples  of  this  kind  it  is  best  for  beginners  to  arrange  all 
the  factors  in  the  denominators  of  the  fractions  so  that  a  precedes 
h  or  c,  and  that  b  precedes  c.  We  therefore  change  b  —  a  into 
—  (^a  —  b),  c  —  a  into  —  (a  —  c),  and  c  —  b  into  —  (&  —  c). 


198 

FRACTIONS. 

Thpn 

1                                1                              -1 

(6-a)(6-c)      -(a-6)(&-c)      (a-&)(&-c)' 

and 

1                                    1                                   1 

(c  -  a)(c  -  b)     {-(a  -  c)}{-(6  -  c)}     (a  -  c)(6  -c) 
Hence  we  have  to  simplify 

1^-1.1 


(a  —  6)  (a  —  c)      (a  -:-&)(&  —  c)      (a  —  c)  (6  —  c) 
The  L.  C.  M.  is  (a  -  6)(a  -  c)(6  -  c)  ;  and 

1  ^   b-c 

{a  —  b)(^a  —  c)      (a  —  b){a  —  c)(b  —  cy 

-1  ^  -(a-c) 

(a-6)(&-c)      (a-6)(6-c)(a-c)' 

and  1  ^  -  ^ 


(a  —  c)  (6  —  c)      (rt  —  6)  (a  —  c)  (6  —  c) 
Hence  we  have 
b  —  c  —  (a  —  c)-\-a  —  b_  b  —  c  —  a  +  c+a  —  b 0_ 


{a  —  b){a  —  c)(b  —  c)     (a— 6)(a— c)(6  — c)    {a  —  b){a—c)(b  —  c) 

EXAMPLES  XL VI. 
Reduce  to  their  lowest  common  denominator : 
1.   A,    ±,     7  g         3  4 


3a;'   5a;'  30a;  '  2a;-2'   3a;-3 

7. 


1         1         1 
2  ax'   6  6x'   8  ex 


^     a      b      c  « 

6c    ca    ab 

.     be    ea    ab  « 

a      6       c 

5.   -i ^—  10. 


5 

3 

6X  +  12' 

1 

8a; +16 
3             5 

a;+l'   2 
6 

x  +  2'    x2-l 
2              4 

5a;-  5' 
a 

3x  +  3'   a;2-l 
X            a2 

a;  +  l'  2  a;  +  2  '   x  -  a'   a  -  a;'  a;2  -  a^'   a^  -  x^ 


11. 

12. 
13. 
14. 


FRACTIONS.  199 

2a  b  3a2  6  6^ 

a-b'  2b -2a    4(a^  -  b^y  6(62 -a^)* 

1  2x  3x2 

a;+  l'    (x+l)2'    {x  +  iy 

1  1  1 


(x  -  a)  (a;  -  &)     (6  -  x)  (c  -  x)  (x  —  c)  (x  —  a) 

1  1  1 

ia-b)(ia-cy   (6_c)(6-a)'  (c-a)(c-6)' 

Reduce  to  one  term : 

IK     „      1  ,      2  2Q  6a-56     4a-76 


iU. 

"      ^-^a  +  l- 

16. 

a  +  x+    ^'    . 
a  -X 

17. 

x-2y 

18. 

a  —  b  .  a  —  6 
a^b        ab^ 

19. 

a-bb     a-Sb 

21     Q^  ~  '^^  I  3q  —  ft 


22.  2^-111-3^  +  ^. 

3  4 

23.  5a;-2y_3y-2x^ 

6  4 

24.  ?  -  ^~^  +  ^  ~  ^ 


4         3  6 


25    2x-3y     x  +  2y     3x-2y 
3  4  6 


Simplify 

26. 

a 

-+        ^ 

a  -  i 

&     6-a 

27. 

X  —  a     a  —  X 

28. 

X 

X2- 

1       ^ 

a2  '  a2  - 

x2 

QQ 

1 

2 

80. 

1 

2  -X 

4 

4-x2 

81. 

1 
3  +  x 

-A 

82. 

1 
x-3 

1 

x-2 

Ait 

1 

1 

1  —  X     1  —  x2  X  —  4yiX  —  5y 


!00 

FRACTIONS. 

34. 

1                   1                     .n       2a     . 

2  6        a2  +  62 

'   Sx-2y      bx-2y                      a  +  ft  ' 

a-h     a'-h'^ 

^iS 

1- 
1- 

^x      \-x                         ^j         a 

^            1 

-X     \-\-x                               a-\ 

a{a  -  1) 

36. 

a+26a-26                   ^^        2 
a-26a  +  2&                         x-2 

1          x  +  6 

x  +  2      x2  +  4 

37. 

1       1         2                         43       1 

1              x  +  3 

X  - 

-  2  '   (x  -  2)2                    ""    x-1     2{x  +  1)     2(x2  +  1) 

38. 

a  - 

a        1        2  a6                   44      ^  ~  ^ 

-2&      (a-2  6)2                  '    1  -3x 

3  +  x       l-16x 
l  +  3x     9x2-1 

39. 

x  +  y      x-y  ^      4.y'^           45        «     + 
X  —  y      X  +  y     x2  —  y2                a  —  ic 

a      ,      2a2 
a  +  X     a2  +  x2 

46. 

a      1      a      1      2a2      ^      4  a* 
a-x     a  +  x     a2  +  x2     a*  +  x* 

47. 

1             11             1     , 

x+1      x+2      x+3     x+4 

48. 

\  ?+  \-         49. 1 

x-lxx+1                         a 

2  +  1  • 

a+1     a+2 

50. 

1            3            3            1 
x-3     x-1     x+1     x+3 

51. 

1         3      1      3             1 

a     a+1     a+2     a+3 

52. 

1             4      1  6         4,1. 

X— 2     x-l'x     x+1     x+2 

53. 

1         4,6            ^      1      ^ 
a     a+1     a+2     a+3     a+4 

54. 

1                       2                       1 
x2-5x  +  6     x2-4x  +  3     x2-3x 

+  2 

55. 

.          1       .    .-     .       .^       _       . 

1 

x2  +  5  ax  +  6  a2     x^  +  4  ax  +  3  a^     x2  +  3  ax  +  2  a2 


FRACTIONS.  201 


gg  x-l 2(x  -  2)  x-S 

(x-2)(x-S)      (x-S)(x-l)      (x-l)(x-2) 


X(X  -  1)        1  -  X2        iC(x  +  1) 

gg  2g x  —  a 2 

(ic-2a)2     a;2-5ax  +  6a2     x-3a* 

59    3  +  2g     2 -3a     16a -q^ 
2 -a        2  +  a        a2-4 

gQ    a;  +  gy     x  -  ay  .  2^^ -^  «^y^ 
x  —  ay     X  -{•  ay        x^  —  a^y^ 

61    a^^-2a;     a;4-3       4x  gg        11         l+2a;-g;2 


68. 


X2-1          X+ll-X  1-XX+l  1 

_J_  +  _«i^^ L_. 

x+3y     x2-9y2     3y-x 


a;^-(y-g)g      y2_(^^a;)2     g2-(x-y)g 

(X  +  y)2  -  Z-2  '^  (y  +  Z)^  _  x2  "^  (;2  +  x)2  -  y^' 

g2_(6  +  c)2       62-(c  +  a)2      c2-(a-6)2 
(a  +  6)2  -  C2'''  (6  -  c)2  -  a2"^  (c  -  a)2  -  ft** 


166.   Multiplication  of  Fractions.     We  have  now  to  show 
how  to  multiply  any  two  algebraical  fractions. 

Let  the  fractions  be  -  and  -• 
b  d 

Let  aj  =  ^X^. 

0     a 


Multiply  by  6  X  d ;  then 


xxb  X d  =  ^x-x bxd 
b     d 


=  ^x6x^xd,       by  Art.  52. 


202  FRACTIONS. 

But  -xb  =  a,  and  -  x  d  =  c ; 

0  a 

.',  xbd  =  ac. 

Divide  by  hd ;  then     x  =  — 

hd 

That  is,  «x-^  =  ^. 

h     d     hd 

Thus  the  product  of  any  two  algebraical  fractions  is 
another  fraction  whose  numerator  is  the  product  of  their 
numerators,  and  whose  denominator  is  the  product  of  their 
denominators. 

The  product  of  any  number  of  fractions  is  found  by 
the  same  rule. 


For  example : 


^x-x-  —  —  x-  =  — • 
h     d     f~hd     f     hdf 


Again,  a     c     d  =  ^ 

0     d     a     hda     o 

And       a^  +  1  ..  a;  +  2  ^  a;  4-  3  ^  (a;  +  l)Ca;  +  2){x  +  3)  ^  ^ 
a  +  2     X  +  3     X  +  1      (ic  +  2)  (x  +  3)  (x  +  1) 


167.   Division  of  Fractions.    Let  -  and  -  be  any  two  f  rac- 
a     c  b         d  ^ 

tions,  and  let  a;  =  -  -r-  -• 
b     d 

mi.  caeca 

Then         a;  x -  =  -^- x -  =  - ; 
d      b     d     d      b 

c      d      ad 
,-.  a;  X  -  X  -  =  -  X  — 
d     c      b     c 


FRACTIONS.  203 


But  X  X-  X-=x\ 

d     c 

a     d 

.'.    X=  -X  — 

b      c 

Thus  to  divide  by  any  fraction  is  the  same  as  to  multiply 
by  its  reciprocal.     [See  definition  of  reciprocal,  Art.  168.] 

Ex.  1.   Divide  -  by  ?• 
a^        a 

a^  '  a     a2     ^     a 

Ex.  2.   Divide  -  by  ab. 
b 

g      .      q^—Qy       1      -.1, 

b  '       ~  b     ab~b^ 

Ex.  3.   Divide  ^-=1^  by  ^i:^'- 
x3  +  a^  x-\-a 

X  —  a    ,  x^  —  a^_x  —  ax  +  a 


X*  4-  a^      x  +  a      x^  +  a^     x^  —  a' 

1 


(x^  -ax+ a2)  (x2 + ax+ a^)      (x*  +  a'^x^  +  a*) 

168.   When  two  quantities  are  such  that  their  product 
is  1,  each  is  said  to  be  the  reciprocal  of  the  other.     Thus 

-  is  the  reciprocal  of  -,  and  -  is  the  reciprocal  of  a. 
a  b  a 

A  quantity  is  small  or  large  according  as  its  reciprocal 

is  large  or  small.     For  example, 

10000  ^j-m^Thi^TJi^  ^y 
but  10000  >  1000  >  100  >  10  >  1. 


204  FRACTIONS. 

If  then  a  be  very  small,  1/a  is  very  large,  and  if  a  be 
very  large,  1/a  is  very  small. 

When  a  becomes  zero,  the  fraction  a/h  assumes  the 
form  0/5  and,  by  the  proposition  of  Art.  127,  is  itself 
zero.  But  when  h  is  zero,  the  fraction  assumes  the  form 
a/0  and,  regarded  as  a  quotient,  has  no  meaning ;  for 
division  by  zero  is  an  impossible  operation.  It  becomes 
merely  symbolic. 

Yet  it  frequently  makes  its  appearance  in  algebraic 
operations,  and  since  it  may  be  regarded  as  having  arisen 
by  virtue  of  a  decrease  of  the  denominator  from  a  finite 
quantity  to  zero,  that  is,  by  an  increase  of  the  fraction 
itself  beyond  measurable  limits,  it  is  called  an  infinite 
quantity,  or  infinity,  and  for  convenience  the  special  sym- 
bol 00  is  used  to  represent  it. 

But  this  nomenclature  must  be  regarded  as  conven- 
tional, and  the  symbol  oo  must  be  used  in  algebraic 
operations  with  extreme  caution.  A  complete  study  of 
its  legitimate  use  requires  more  extended  discussion  than 
is  at  present  appropriate.  [See  Treatise  on  Algebra, 
Art.  217.] 


3.    -H- 
4. 


EXAMPLES    XLVii. 

duce 

the  following  fractions  to  their  simplest  f( 

2a, 
3c  ' 

4a 

5. 

h     d     ac 

5a2 
6  6c 

X   ^^' 
10  ca 

6. 

«'  X  ^'  X  c2 

be     ca     ab 

« 

c 

7. 

b^b^c^ 

c 

d 

a     c  '  a 

2a2 
6c 

3a& 

8. 

a\b^  .  c2 
b^     c^'  a^ 

FRACTIONS. 

9.  ?«x2»xi«. 

be      ca      ab 

5  62       10  62a; 

10.   2x»      3,3      5^ 
3yz     bzx     2xy 

,„.    5a26c  .  5a26c2 
3  62c2a  ■  3a62c» 

205 


13.  By  what  must  -^  be  multiplied,  that  the  product  may  be  —  ? 

5  x^y  5  X 

14.  By  what  must  ^-^  be  divided,  that  the  quotient  may  be  —  ? 

Express  in  their  simplest  forms : 

16.   ^-iJLx-^+iL.  21.   ^^x^^x^^. 

x^  -\-zy     xy  —  y^  x  —  2     x-S     a:-4 


16. 


g  +  6    ^  q6  -  6^  gg       x+1    ^   x+2         x+3 

a8-a26      a6  +  a2  '    (a;+2)2     (x+3)2     (x+l)2* 


j^    x2  +  2 X  ^  z2  _  3 a;  ^^    x2-3x  +  2^^x2-7x+12 


a;2_9        x2-4  x2-5x+6     x2-5x+4 

18      x2-y2   ^x-2y.  ^^          x2-l       ^^     x2-25 

*  x2-4y2       x-^y  *  x2+3x-10    x2-3x-4 

19.   ^  +  3a;2 ^   x  +  3  g^  a^  -  a^  ^  x  +  2a_ 

X  +  4     *  x2  4-  4  X  *  x2  —  4  a2       x  -  a 

20      <^  +  ^^    ^  q6  +  4  62  gg  g^  -  x»  ^  (g  -  x)2. 

g2  +  6  g6  ■  g8  +  5  g26'  *  g*  +  x^  '    g2  -  x2  * 

87.      g  +  ^    X  ^^  ~  ^^  '   g*  -  X* 


(g  -  x)2     g2  +  a;2      (a  +  x)8 

(g-6)8         g*-6*     "    g2  +  62  * 
^     (g  -  6)2  -  c2  ^  62-  (c  -  g)2 


80. 
31. 
32. 


(g-c)2-62'^c2-(g-6)2 

x^-(y  +  zy  ,  y^-(x  +  zy 
x-^-Cy-zy  •  yi-{x-z)^' 
ofi  +  y^  X  '^~y  ^  a;*  -  x2y2  +  y* 

X^  —  y^        X  -\-  y        X*  +  x2?/2  -j.  y4 


206 


FRACTIONS. 


169.   We  will  now  give  some  examples  of  more  com- 
plex fractional  expressions. 


d     ad 
'be 


Ex.  1.  Simplify  ^  /-- 
hi  d 

a / c_a     d 

Ex.2:  Simplify  (I +iy(^  +  l). 

\b       1/  [a       I         b    I     a  b         b  +  a     b 

Ex.3.    Simplify    (^-^ -^^^Wf  «±^+ «II^V 
\a  —  X     a  +  xj/   \a  —  X     a  +  xj 

a  +  x     a  —  X  _  (a  +  x)  (a  +  x)  —  (a  —  X)  {a  —  x)  _    4:  ax 
a  —  x     a  +  x~  (a  —  x)(^a  +  x)  ~  a'^  —  a;^' 


a2-x2 


a-\-xa-x_(a  +  x)(a  +  x)  +  (a  —  x)(a-  x)  _  2a^  +  2x^ 

2L11Q.  —  -|-  ; —  7- r-7 ; r ^_— — -^-— 

a  —  X     a-\-  X  (a  —  ic)  (a  +  a;) 

Hence  the  given  fraction  is  equal  to 
4  ax     /  2  a2  +  2  a:2 


4:  ax     /2a; 
a^-x^l     a- 


Ex.  4.    Simplify 


x  +  2 

X 


x  +  1  (x  +  2)  ic  -  (x  +  1) 

X 

a;(a;2  +  a;-l) 

~  X-    ^'^  +  ^^    ~  ^(^^  +  X  -  1)  -  (x2  +  2x) 

X2  +  X  -  1 
_  X(X2  +  X  -  1)  _  X2  +  X  -  1. 

x8  -  3»  x2  -  3     ' 


FRACTIONS.  207 


a  —  b   ,    b  —  c 


T^      E      o-      ^■e^      I  +  ab     1  +  bc 

Ex.  6.     Simplify ; t—tt r— 

1        (a-b)(b-c) 

(l  +  a6)(l  +  &c) 
a-b       b-  c  _  (a  -  6)  (1  +  be)  +  (6  -  c)  (1  +  ab) 
l  +  ab     l  +  bc  (1  +  ab)  (1  -^  be)      ^ 

_a  —  b  -{-  abc  —  b'^c  +  b  —  c -{■  ab^  —  abc 

il  +  ab)(l  +  bc) 
_a-c  +  ab'^-cb^ 
(l  +  a6)(l  +  6c)' 
^        (a-b)(b-c)    ^(l  +  a6)(l  +  &c)-(a-6)(6-c) 
(1  +  a6)(l  -f  be)  (1  +  a6)(l  +  be) 

_  1  4-  a6  +  6c  +  abH  -  ab  -\- b^  -\-  ae  -  be 
(l  +  ab){l  +  bc) 

_  1  +  gc  +  fe^  +  ab^e 
(l  +  ab)il  +  bc)' 

Hence  the  given  fraction  is  equal  to 

a  -  e  +  ab^  -  eb"^  .  1  +  ac  +  6^  +  ab^c 
(l  +  a6)(l  +  6c)  ■    (l-hab)(l  +  be) 

_  a-c  +  ab^-cb^  __  (g  -  c)(l  -f-  &^)  _  a-c 
1  +  ac  +  62  +  acb^     (1  +  ae)  (1  +  62)      i  ^.  ^c 

EXAMPLES  XLVIII. 

Ig— 2x     X  — 2g/      \2g— x      a  —  xi 

2.    [     ^       I   ^~^]    '    f     ^ 1-^]  ■ 

*\l  +  x         x/ll+x         xJ 

4.      2     ^      ^ 


Simplify 


1_1     i_y      i_? 
X     ^  X  y 


208  FRACTIONS. 


X  y  2  1 4-  a;      1  +  a;^ 


1+?  1+r  1+1         10.  1  +  ^^  i  +  "l 

^y        ^x     x^y  1  +  x^  1  +  x« 

1  +  x3  1  +  x* 

^4.2^  +  2     ?  +  ?'-2  ,,     1  1 


X  y     X  11. 


6.  ^__ri_4-2 •"  "•   ^ T 

x  +  y  x-2/  x+ 


3.      1 


a+&  _^6  » 


a +  26      a  12.    x3-         ^ 


a  +  6  &  ^      1  -X 

a        a+26  ^  ,  1 


X  +  1  13. 


« r^-  .+     1 


o  —  a 

x  +  2  _ 4x  +  5  j^                1 

2x  +  3      5x  +  6  *    ^2        x«  +  1    ' 

2x  +  3  _  3x+4'  ^  _j_      1 

3x  +  4     4x+5  x-1 


170.   The  following  theorems  are  of  importance. 

Theorem  I.    If  the  frdctions  a/h,  c/d,  e//,  etc.,  he  all  equal 
to  one  another,  then  will  each  he  equal  to  the  fraction 

pa -{•  qc -\- re -\ 


ph-\-qd  +  rf-\-'" 
Let  each  of  the  equal  fractions  be  equal  to  x. 
Then,  since  a/h  =  x,  c/d  =  x,  e/f=  x,  etc., 

.'.  a=bx,    c  =  dx,    e=fx,  etc.; 

.',  pa  =phx,   qc  =  qdx,    re  =  rfx,  etc. 


FRACTIONS.  209 

Hence,  by  addition, 

pa-^qc  +  re  -\ =pbx  +  qdx  +  rfx  -\ 

=  (pb-i-qd-\-rf+'")x', 
pa -\- qc -{•  re -i-  '  • '  __     _  a  _ 

As  a  particular  case,  each  of  the  equal  fractions  a/&,  c/d,  c//, 
etc.,isequalto"  +  ^  +  ^  +  '' 


6  +  d+/+- 


Theorem  II.     If  the  denominators  of  the  fractions  a/h, 
c/dj    e/f    etc.f   be    all   positive,    then    will    the   fraction 

—  be  greater  than  the  least,  and  less  than  the 


a+  c  +  e4- 


b-i-d+f  + 

greatest,  of  the  fractions  a/b,  c/d,  e/f,  etc. 

Let  a/b  be   the   greatest   of    the   fractions,   and  let 
a/b  =  X. 

Then  c/d  <  x,  e/f<  x,  etc. 

Hence,  as  b,  d,f"-  are  all  positive,  we  have 

a=bx,  c<dx,  e<fx,  etc. 

Hence,  by  addition, 

a-\-c  +  e-{-'"  <bx  +  dx-\-fx-{-  ...  <  (6  +  ^+/+ ...)a;; 

a  -\-c  +  e  -\- 


b  +  d-\-f+ 


<x. 


a  I   c   I   6   I  , , , 
Hence  —^ — — — — —  is  less  than  the  greatest  of  the 
b-hd+f-h'" 

fractions ;  and  it  can  be  similarly  proved  to  be  greater 

than  the  least  of  the  fractions. 


210  FRACTIONS. 


Ex.  1.   Show  that,  if  ^  =  ^,   then  will  ^i^  =  ^-±-^. 
b     d  a—bc—d 

Let  ?  =  X ;  then  -  =  x.    Hence  a  =  bx,  and  c  =  dx. 
0  d 


Hence 
Also 


a  +  b  _ 
a-b 

_bx+b  _ 
bx-b 

_x  +  \ 
x-1 

c-\-d  _ 

_dx  +  d  _ 

.x  +  1, 

dx  —  d      X  —  \ 


And,  since  ^-i-?  and  ^-+-^  are  both  equal  to  ^  "*"    ,  they  must 
a—b  c—d  x—1 

be  equal  to  one  another. 

Ex.  2.    Show  that,  if  ^  =  ^,   then  will  ^!-+_^ 
b      d  "  ■    ^' 


Let|  = 

=  x; 

then  ^  =  X.     Hence  a  =  6x,  and  c  =  dx. 
d 

Hence 
Also 

a2  +  62  _  52a;2  +  62  _  62(a;2  +  1)  _  ^2 
C2  +  ^2       ^2^.2  +  ^2       ^(a;2  +  1)       d2 

a2  _  62  _  62x2  -  62  _  62(x2  -  1)  _  62 
c^-d^     d^x^-d^     d2(x2-l)     d2 

Hence 

a2  +  62  _  a2  _  62 

C2  +  (?2        c2  -  d2 

Ex.  3. 

Show  that,  if  ^^  +  ^^  =  «^  +  ^^  =  ^^  +  «2^, 

in  will 

bcx            _          cay          _          a60 

—  al  +  bm  +  en     al  —  bm  +  en     al -h  bm  —  en 

By  Theorem  I.,  each  of  the  given  equal  fractions  is  equal  to 

—  a(cy  +  bz)  +  b(az  +  ex)  +  e(bx  +  ay)  _  2  6cx 

—  al  +  bm  +  en  —  al  +  bm  -\-  en 

And  similarly  each  of  the  fractions  can  be  proved  to  be  equal 

to  2cay ^^  ^^   2abz 

al  —  bm  +  en  al  +  bm  —  en 


FRACTIONS. 


211 


Simplify 

24  a66c8x2* 


EXAMPLES   XLIX. 


4. 


x'^-Qxy  -ifby^ 
«*  —  6  x^y  +  4  xy2 

a^-lOx^y-  lla;y2 
«2y  _8a;y2_9y8  ' 


6. 


7. 


9. 


(a  -  6)  (a*  -  h^) 

(q2_62)(a6_7)6) 

(a  +  h)\a  -  6)2  ' 

ac8  +  3x2-x-3 
x*  +  4x8-  12x-9* 

3a^+5a^-7a;2  +  2a;  +  2 
2x*+3ic8-2a;2  +  12a;  +  5' 


IQ    2a:*  +  9x8y  +  14a;y8  +  3y* 
3ic*  +  14ic8y  +  9a;y8  +  22/** 


11. 


2  a^  -  11  a;2y  -f  11  a;y2  +  4  ys 


12.   2^-±-i^-3^^ 
3  5 


2x*-3x8y  +  7x2 


12  xy^  —  4  y* 

a 


13. 


14. 


16. 


16. 


^x-2y     ^y-2x 
4  6 


4-a; 

1 


X2  -16 
3     . 


S-x     a;2 
a; 


9 


X  -  2  y      (2  y  -  x)2 


17 


18. 


19. 


20. 


21. 


+ 


4ab 


a -4b      (4  6-a)2 

_1 1_  + J^ L_. 

a+4    a  +  6    a  +  6     a  +  T* 


+ 


a+3      a+6     a+9 


x3 


?/" 


x2-9y2 

X2_v2 


x  +  3y 


y-^. 


x2_4y2     2y-x 


22.  |/^l  +  lV-J--_A_UI/'i-iV-i-V 

Ha     6/       ab      {a-byi   '  \\a     b)       ab ) 

23.    (     ^       I       Mxf^-^ ^V 

Vx-2      x-8/     V3x-8     x  +  2y 


(x— y)*— xy(x— y)2— 2x2y2 
(x-y)(x8-y3)+2x2y2 


25. 


I      2x  +  l 

1        X2  +  X  +  1 


212  FRACTIONS. 

««  1         .         1  3 


27. 


6a-26     3a  +  26     6a  +  2b 
1  1.6 


8a  +  26      2&-8«      16a2-62 
a6  +  &6  ^   a-&  ^  gi  -  ^252  4,  54 
a6  _  56     ^2  +  62  •  a;4  +  ^262  +  6** 

m2+  w2 

n  m2  —  n2 

11  m^  —  n^ 

n     m 

/l  +  ^  +  _^i_\|l ?i_l 

I        a  +  6      (a+6)2J    I        (a  +  6)2J 

I         (a  +  6)3i   I        a  +  6i 


31.   x  +  '</     x-y        4 y2 
'   X  —  y     x  +  y     x^  —  y^ 


32 
33. 


I        a(a-36)i  I       26-ai 


xyiy^  —  a;2)      x2  +  ic?/     xy  —  y^ 


34.     a+6  ^   a2-62 6-a.       35      ^-^-  ^^ 


a+6  6    25  62-a2     a -56  a-3  6     a+3  6     9  62-a2 

36.   ^ "^^^ +  1        ^-^ 


2(x-l)      x2-7a;+10     2cc2-9a;+18 
37.    ^ 2 x-6 33       3  2  -.-7 


a;_3    a;_4     (x-2)(x-5)  x-4    x-S     (x-6)(x-3) 

39.  (x-y 1 lx?i+^. 

I  x-y  J 

40.  fa^  +  y i 1  x^'-y'. 

I  X  +  7fJ 


FRACTIONS.  213 

"•     l^       1  +  x^    l-x2  /''2X  +  1 


a;-2 ^     x-4: 1 

42.  ?-2 


X 


a;_2 ^     x-4  ^ 


X  —  5  X  —  4 

43.   .6_^__7_^^  1 


x2  +  2x-8      X2  4-X-12      x2-5x  +  6 
44     rxM  v«^f'^'  +  ^'     ^'-y'U  ^  •  l^+y     ^-y] 

^  ^1x2-2/2        a;2+«/2J        ^2  -  Xy  +  J/^    '    \  jg-y       x-^y  ( 

V    2a6  ja8  +  68  ■  a2  -  a6 -f  62* 

47.    I 1 + 1^ U/_^±iL_^y_l. 

1 3x2- 14  xy+ 15  2/2     3x2-2  XI/-5  2/2/      lx-3y      x+Syi 

48     ^  ~  ^  I  c  —  g  ,  g  —  6  ,  (6  —  c)(c  —  a)Ca  —  6) 
*   a  +  x      6  +  x      c  4- a;      (a  +  x)(6  +  x)(c  +  x) 

I.\.a  +  ft/        a  +  6        i       lU-fc/         u-b        I 
61.       16  21  16  21 


17     x-9     x  +  4     x  +  7 
11  7  11  7 


x-6     x-7     x+2     x+4 

x  +  7x  +  8     x  +  9     x+10 
x+1     x+2     x+3      x+4 


214  FRACTIONS. 

{a  +  h-cY-d^         (b  +  c-ay-d^         (c  +  g-by-d'^ 
(a  +  by  -  (c  +  dy  "^  (6  +  cy  -  (a  +  dy  "^  {c+ay  -  ib  +  dy' 

55.   Simplify    ^  1.1 


axy     aQii^  a)  {y  -  a)     y{x  -  a){y  -  a) 

56.    Simplify ^+-^ + ^±^ + ^-±-^ 

xix-y){x-z)     y{y  -  z){y  ~x)     z{z-x){z-y) 


ab 


57.  Find  the  value  of  - — —  when  x  = 

b  -2x  a  +  b 

58.  Find  the  value  of 

a  +  b  +  2c     a  +  b  +  2d^ 
a  +  b  -2g     a  +  b  -2d 

when  a+b  =  -i^- 

c  +  d 


59.   If  -  =  -,  prove 
b     d 


(i-) 
(ii.) 


a 

-b 

c-d 

n 

a  - 

-2b 

c~2d 

a2 

c2 

(c  +  dy 

(i 

(a 

+  6)2 

f 

X 

.    y    _ 

z 

gg  _  52  ^  C^  -  ^2 

a2  +  62       c2  +  d2* 

jpa2  4-  qah  +  rb^  _  pc"^ + qcd  +  rd^ _       ,: 
Za2  +  ma&  +  nbf^      Id^^  mod  +  wd2 


go.   If  -^^L-  =  —y—  =  _£_  ;  then  will  x  +  y  +  z  =  (i. 
a  —  b     b  —  c     c  —  a 


Gl.   If     y+^  __gjL^^_Ml_  .  then  will  each  fraction  =     ^ 
ay+bz    az  +  bx    ax+by 

62.    Prove  that,  if  a,  b,  c  be  unequal,  and 


ay+bz    az  +  bx    ax+by  a+b 

b  —  c     c  —  a     a  —  b 


X  y  z 

then  will  x  +  y  +  z  =  0  and  ax  +  by  +  cz  =  0. 

63.   Prove  that,  if  5Li|  =  J-tS.  =  ^iL«^, 
a  -  6     2(6  -  c)      3(c  -  a) 

then  8a  +  9&  +  6c  =  0. 


FRACTIONS.  ^15 

64.   Prove  that,  if  ^^  -  ^V  =  ^x-az^  ^^^^^  ^^^^  .^  ^^^^  ^^  ay-bx^      - 
b  —  c         c—a  a  —  b 


65. 

Prove  that, 

if 

a 

b 

c 

2y+  2z- 

-Sx 

2z 

'.  +  2x- 

-Sy 

2x 

+  2y- 

-Sz 

en 

X 

_ 

y 

z 

a  +  2b  +  2c     b  +  2c  +  2a     c  +  2a  +  26 

66.   Prove  that,  if 

x  —  a  _y  —  b  _z  —  c 
P  Q     ~     r    * 

and  x+y  +  z=ia  +  b  +  Cj 

then  x  =  a,  y  =  b,  and  z  =  c. 

g^    jf  ^y±cz^cz±ax^qx±by^2^^ 

a  b  c 

prove  that  cix       ^       by       ^       cz       ^^^ 

b  +  c  —  a     c-\-a  —  b     a  +  b  —  c 

and  therefore  x=bc(b  +  c  —  d)^ 

y  =  ca{c  +  a  —  b)y 

M  =  ab  (^a  •{■  b  —  c). 


216  EQUATIONS   WITH  FRACTIONS. 


CHAPTER  XV. 

Equations  with  Fractions. 

171.  In  the  present  chapter  we  shall  give  examples  of 
equations  which  contain  fractional  expressions. 

Ex.1.     Solve    ^^_3^-1^43-5:«. 

^  ^^  ^^  % 

We  may  multiply  both  sides  of  the  equation  by  120,  the  L.  C.  M. 
of  the  denominators  of  the  fractions,  without  destroying  the 
equality  ;  we  thus  get  rid  of  fractions,  and  have 

24(x  -  1)  -  15  (3x  -  1)  =  20  (43  -  5x)  ; 
...  24a;  -  24  -  45x  +  15  =  860  -  lOOx. 
Transposing,  we  have 

24x  -  45x  +  100a;  =  860  +  24  -  15 ; 
.-.  79  a;  =  869. 

TT  869       n  1 

Hence  a;  =  ^^^  =  11. 

79 

Ex.  2.     Solve  — ^  + 


a;-3      x  +  3     a;+5 

The  L.  C.  M.  of  the  denominators  is  (x  —  3)(a;  +  3)(x  +  5). 
Multiplying  both  sides  of  the  equation  by  the  L.  C.  M.,*  we  get 
rid  of  fractions,  and  have 

(a;  +  3)(a;  4-  5)+  2  (a;  -  3)(x  +  5)=  3  (a;  -  3)(a;  +  3); 
.-.  a;2  +  8x  +  15  +  2a;2  +  4x-30  =  3a;2-27. 

*  See  Art.  181. 


EQUATIONS   WITH  FRACTIONS.  217 

By  transposition,  we  have 

a;2  +  2x2  -  3x2  +  8x  +  4a;  =  -  27  -  15  +  30 ; 
.-.  12x  =  -12, 
or  x  =  —  l. 

Ex.3.     Solve  1      .      1     _     1      .      i 


x+ 1     x+7     x+S     x+5 

In  this  case  it  is  best  not  to  multiply  at  once  by  the  L.  C.  M.  of 
the  denominators ;  the  work  is  simplified  by  proceeding  as  under. 

Wehave  — ^  +  — ^^ ^  =  0; 

x+1     x-^S     x+7     x+5 

x  +  3-(x+l)x-f5-(x  +  7)_ 
••    (x+l)(x  +  3)       (x  +  5)(x  +  7)  -"' 

that  is =  0 ; 

(x+l)(x  +  3)      (x  +  5)(x  +  7) 

,   .-.  (x+l)(x  +  3)  =  (x  +  5)(x  +  7), 
that  is  x2  +  4x  +  3  =  x2+ 12X  +  35; 

.-.  4x-12x  =  35-3; 
.-.   -8x  =  32, 

.-.  x  =  32/(-8)  =  -4. 

Ex.  4.    Solve  the  equation 

X  —  1  ,  x+5 _  x  +  1     x  +  3 
x+1     x+7~x+3     x+5 

x+5_j_     2        x+l_i         2 


Since - 

X 

-1 

+  1 

:1- 

and 

wehave 

1-- 

2 

x+l'x  +  7  x  +  7'x  +  3  x  +  3' 

x_+3_  j__2_ 
x  +  5  x  +  5' 


x  +  1  x+7  x  +  3  x  +  6* 

2  2^2  2 


x+1     x+7     x+3     x+5 
which  is  the  same  equation  as  that  in  Ex.  3. 


218  EQUATIONS   WITH  FRACTIONS. 

Ex.  5.   Solve  the  equation 

4x+5  ,  a;+ 5_2x +  5     a;'^  -  10 
x-\-l       x  +  4       x  +  2        x  +  3 

By  division,  we  have 

4a;  +  5_^  J       1        a;  +  5  _  .^         1 


a;+l  jc4-lic  +  4  x  +  4 

ic  +  2  a;  +  2  x  +  3  x  +  3 

Hence  4  +  -1-+ 1 +-l-  =  2  +  -l-- fx-3--^V^ 

x  +  1  x  +  4  x  +  2     V  x  +  SJ 

■       1     +     1-1       ■       1 


x+1      x+4     x+2     x+3 

.    _J 1_^_J. 1_. 

**x+l      x  +  2      x  +  3     x  +  4' 

(X  +  2)  -  (X  +  1)  ^  (X  +  4)  -  (X  +  3) 
(x  +  l)(x  +  2)  (x  +  4)(x  +  3)  ' 


that  is 


(x  +  l)(x  +  2)      (x  +  4)(x  +  3) 
.-.  (x+l)(x  +  2)  =  (x  +  4)(x  +  3), 
that  is  x2  +  3x  +  2  =  x2  +  7  X  +  12  ; 

.'.  x2  +  3x-x2-7x=12-2; 
.-.   -4x  =  10; 

10  5 


2 


EXAMPLES  L. 


1    a?  — 2     X  —  3_x— 7  „    X     x  —  2_x     o^  —  3 

6  4~10'  *6         5~5  4* 

o3x  — 1      oa^-l_2  — X  ^    x-2    gX-l,.       ^x-2 


11  6  10  3  5  6 


EQUATIONS   WITH   FRACTIONS.  219 


g    l-3a;     3x  +  l^      2  ^    S^ 5_J ]_  ^  ^ 

2  2  l-3x'  *   2x  +  2      4ic  +  3     4x 

g    3-4a;     5-8x_l-a;  ^^    2x-  5_2x-7 
6               12         1  +x  '    3x-7      3x-5* 

„    Sx-1     o«+l_o     3-6x  „„    6x-2      3x+r 

7.  — z- — -—z —  ad. -  = 

4  X  — 3  8 

8    2x     3^  —  1  —  3      1— 4x  rto 


6         X  +  1  10 

1.1  2 


8x 

-5 

4x+8 

X- 

-9 

2x- 

-5 

3x 

-7 

6x- 

-4 

3- 

2x 

2x- 

-7^ 

10. 


4x  +  6     6x  +  4     2x  +  3  x-5        4-x 

1       I        2^2.         gg    a;+l     x-3^8 
3x  +  9     5x+l      a;  +  3  'x-1     x4-3     x 


J  J     ^^ 3      _      4  gg    x  +  2     a;-2_     8 

4x+12      4x+l      2x+6*  'x-2     x  +  2~x+l' 

12.  _A_  +  _5_  =  _J_.  27.   ^^-^±2  +  10  =  0. 
2x  +  3     4x  +  6     6x  +  8  x+3     x-4      x 

13.  _i^ ^  =  3.  28.  -^-3^^+     ^     =0. 

x+lx-2  x  +  2        x-2     x+1 

14.  ^ ^  =  2.  29.   3^^  +  2^+1  =  5. 

X  +  6X  +  6  x  +  1        x-1 

15.  _6x x_^5^  3Q    5^^:i2_2X^^3^ 

X-7X-6  x+2        x+3 

16.  ^^ i^+2=0.  31.    -l^  =  -i L.. 

X  +  3X+7  x3-lx-4x  +  4 

17^       1      I       2_3  o«12x,2_2 


18. 


x  +  4     x  +  6     x  +  5  x3-9     x  +  3      x-3 

_3 2_^^_.  33         4  1^1 

x+1     x  +  2     x  +  3  *x2-l     x+1      1-x 


19.   ^-i §-J_  =  -i_.         34.        ^       •       1     -     2 


2x  +  4     2x  +  2     x  +  6  x2-9     x  +  3     3 


220  EQUATIONS   WITH  FRACTIONS. 

OK    2a;  +  l  8  2a^-l 


2a^-l      4x2-1      l  +  2aj 

36. 

3x  +  5           5       _8  +  3cc 
3x-l      1-9x2     l  +  3x* 

?7 

1              3      _             5 

x-5     2x-6      (x-3)(x-6) 

38 

1              3     _        7x+l 

2x-4     x-5      (x-2)(x-6) 

39. 

^      +     1^      +      4      =0. 
2-3x     6-6x      10  +  x 

40. 

«       1        ^              1      -0. 
3x-l      3-7x     5  +  x 

41. 

1111 

x+5     x+6     x+6     x+8 

42. 

1111 

x+7     x+1     x-\-l     x  +  3 

43. 

1     +      1      _     1     +     1 
x  +  2      x+10     x  +  4     x  +  8 

44. 

1111 
x-2     x-6     x-4     x-8 

45. 

1     +     1     =     ^     +     1. 
x-5     x+2     x-4     x+1 

46. 

X         x-9_x+lx-8 

X  —  2      x-7     X  —  1     X  — 6 

47.  2x+l  ^  2x  +  9^2x  +  3     2x  +  7_ 
*x+l        x  +  5        x  +  2        x  +  4 

2x-3      2x-4      2x-7      2x-8 


48. 


2x-4     2x-5     2x-8     2x-9 


49    »^  +  7     x  +  9  _  x  +  6  ^  x  +  10 
x+6     x+7     x+4'x+8' 


EQUATIONS  WITH  FKACTIONS.  221 

gQ    16  a;  -  13     40  x  -  43  ^  32  a;  -  30     20  x  -  24 
'     4x-3  8x-9         8x-7         4x-5* 

51    a;-7     x-8^2x-7     2x-ll 
x-5     x-6~2x-5      2x-9* 


7  11  7  11 


53. 


x-9     x-4     x+2     x+3 

16     _    21     ^    16    _    21 
x-17     x-9     x  +  4     x  +  7* 

^    x-\-a  j  a^  +  ^_2. 
*   X—  6     X  —  a 

55    ^  +  <^  _  X  — Z)  _  2(g  +  6) 
'   x—a     x+b  X 

-fl       ax     .     6x        ^  ,  7, 
66. 1 =  a  +  0. 

a  +  x     6  +  X 

--      q  +  c    ,    6  +  c  _  q  +  6  +  2c 
'   x  +  26     x  +  2a      x  +  a  +  6  * 

68    ''^~  ^  _  ^  —  <"'  —  2(a  —  6) 
x  —  a     X  —  b     X  —  a  —  b 

172.   Simultaneous  Equations.     Any  pair  of  equations  of 

the  first  degree  in  two  unknown  quantities  can  be  reduced 

to  the  form 

ax-{-by  +  c  =  Of 

a'X'{-b'y-{-c'=0, 

and  any  formula  that  completely  solves  them  solves  also 
every  pair  of  equations  of  this  class.  We  shall  now 
derive  such  a  formula,  applying  for  this  purpose  the 
method  of  elimination  by  undetermined  multipliers. 
[Art.  106.] 

To  solve  the  simultaneous  equations 
ax  +  &?/  +  c  =  0, 
a'x  +  b'y  +  c'  =  0. 


222  EQUATIONS   WITH  FRACTIONS. 

Add  the  first  equation  to  k  times  the  second,  thus  : 

(a  +  ka')x  +  (6  +  kh')y  +  c  +  A:c'  =  0,      .     .    .     (i.) 
and  make  h  +  kh'  =  Q, 

for  the  purpose  of  eliminating  y.    Then 

k  =  -^, 
b' 

and,  substituting  this  value  of  k  in  equation  (i.),  we  obtain 


(«-^«') 


x-\.c--'C'  =  0 
h' 


whence  x  =(  —  c  +  —  •  c' j /(  a 1  '  ^'\ 

Again,  for  the  purpose  of  eUminating  x,  make 
^  a^ka'  =  0. 

Then  lc  =  -^^ 

a 

and,  substituting  this  value  of  k  in  equation  (i.),  we  obtain 

\        a'      I  a' 

whence  y  =  (-c  +  ^^- c<\/ lb -^'h'\' 

When  simplified,  these  fractional  expressions  for  the  values  of 
X  and  y  have  the  form 

x  =  (bc>  -b'c)/{ab'  -a'b), 

y  =(ca'  —c'd)/{ab'  —  a'b). 

Observe  that  either  one  of  them  may  be  derived  from  the  other 
by  interchanging  a  with  b  and  a'  with  b'.  Thus  the  principle  of 
symmetry  here  finds  an  application  to  elimination ;  and  it  fre- 
quently happens  that  by  taking  advantage  of  this  principle,  one-half 
or  two-thirds  of  the  work  of  elimination  by  the  ordinary  process  is 
avoided. 


EQUATIONS   WITH   FRACTIONS.  223 

173.  A  convenient  device  in  elimination,  sometimes 
called  the  rule  of  cross-multiplioation,  is  suggested  by  the 
cyclo-symmetry  observable  in  the  equations 

y  =  (ca'-c'a)/(ab'-a'b)y 
when  they  are  written  in  the  form 

y 


6c'  —  b'c     ab'  —  a'6'    ca'  —  c'a     ab'  —  a'b' 
or  more  concisely,  in  the  form 

a;       _       y        _        1 


be'  —  b'c     ca'  —  c'a     ab'  —  a'b 

For  the  purpose  of  making  the  symmetry  of  the 
expressions  more  apparent,  c  is  changed  to  cz  and  c'  to 
c'z,  in  the  two  equations  to  be  operated  upon,  which  then 
become 

a>x  -{■  by  -\-  cz  =  0, 

a'x  4-  b'y  +  c'z  =  0 ; 

and  the  process  of  elimination  (in  any  of  its  forms) 
gives,  not  now  the  values  of  x  and  2/,,but  of  the  fractions 
x/z  and  y/z,  in  the  form  already  written  above,  namely, 

a?       _       y       _       2 


be'  —  b'c     ca'  —  c'a     ab'  —  a'b 

in  which  z  has  now  taken  the  place  of  1  in  the  third 
numerator. 

The  rule  of  elimination  by  cross-multiplication  may 
now  be  stated  as  follows :  Write  three  fractions  with  x, 
y,  and  z  for  their  respective  numerators.  Under  x  write 
the   difference   of   cross-products  be'  —  b'c,  formed  from 


224 


EQUATIONS   WITH  FRACTIONS. 


the  coefficients  of  the  terms  that  do  not  contain  a;,  and 
under  y  and  z  respectively  write  the  similar  differences 
ca'  —  c^a  and  ab'  —  a'b,  observing  the  cyclic  order  in  the 
displacement  of  the  letters.  By  this  simple  rule  the 
values  of  x/z  and  y/z  are  written  down  by  inspection. 

These  cross-product  differences  are  sometimes  exhibited 
in  the  form  of  square  matrices,  which  are  called  determi- 
nants, thus : 


b 

c 

c 

a 

a 

b 

b' 

c' 

> 

c' 

a' 

J 

a' 

b' 

and  when  so  written  they  suggest  at  once  the  rule  of 
cross-multiplication.     For  example, 

^      -     =bc'-  b'c, 

and  the  other  two  similar  identities  are  derived  from 
this  one  by  a  cyclic  displacement  of  the  letters. 

The  method  here  used  is  applicable  to  the  solution  of 
simultaneous  equations  with  any  number  of  unknown 
quantities.     [See  Treatise  on  Algebra,  Arts.  145,  432.] 

Ex.  1.  Write,  by  the  rule  of  cross-multiplication,  the  values  of 
x/z  and  y/z  which  satisfy  the  simultaneous  equations 

Sx  +  6y-\-2z^0, 

2x  +  1y  +  4z  =  0. 

Solution 


or 


X 

y 

= 

z 

20 

- 

14 

4- 

12 

21- 

lo' 

X  _ 
6~ 

y 

-8 

= 

z 

11 

) 

•• 

X  _ 

z 

6 

y 

z 

=  - 

8^ 
"ll* 

EQUATIONS   WITH  FRACTIONS.  225 

Ex.  2.  Write,  by  the  rule  of  cross-multiplication,  the  values  of 
x/z  and  y/z  which  satisfy  the  simultaneous  equations 

2a;  +  72/-42!  =  0. 


Solution : 


_      y     - 


or 


20-14     4  +  12     21  + 10 

?  =  ^  =  — • 
6      16     31 ' 

'*'  z~Z\    z~Vl 

Ex.  3.  Determine  the  relations  that  must  exist  between  the 
coefficients  of  ax^  +  ftx  +  c  and  a'x^  +  6'x  +  c',  in  order  that  the 
equations 

ax2  +  6x  +  c  =  0, 

a'x2  +  6'x  +  c'  =  0, 

may  have  a  common  root. 

By  a  common  root  is  meant  a  value  of  x  which  satisfies  both 
equations.  Hence,  if  we  assign  to  x,  for  the  moment,  the  particular 
value  in  question,  the  two  equations  become  simultaneous  in  x 
and  x^. 

Then,  by  the  rule  of  cross-multiplication,  we  have 


6c'  —  6'c     (M^  —  da     ah'  —  a'b  ' 

and,  placing  the  product  of  the  first  and  third  of  these  fractions 
equal  to  the  square  of  the  second,  we  obtain 


(be'  -  b'c)  {ab'  -  a'b)  {ca'  -  c'a)'^ 
whence,  after  dividing  by  x^  and  inverting, 

(be'  -  b'c)(ab'  -  a'b)  =  (ca'  -  c'aY, 
which  is  the  relation  sought. 


226  EQUATIONS   WITH   FRACTIONS. 

174.  When  a  set  of  equations  can  be  conveniently 
written  in  terms  of  a  series  of  equal  fractions,  Theorem 
I.  of  page  208  provides  still  another  useful  method  of 
elimination. 

Ex.  1.  Solve  the  simultaneous  equations 

y  +  z  _  z  +  x  _  x  +  y  _  j^ 
a  b  c  ' 

By  Theorem  I.,  page  208,  we  have 

—  y  -  z  +  z  +  x  +  x  +  y  _         2x         _  j^ 
—  a  +  6  +  c  —  a  +  b  +  c 

.'.  x  =  lk(-a  +  b  +  c); 

and  by  a  cyclic  displacement  of  the  letters 

y  =  lkia-b  +  c), 

z  =  ^k{a  +  b  -  c). 

Ex.  2.  Solve  the  simultaneous  equations 

a  b  c 

By  Theorem  I.,  page  208,  each  of  these  fractions  is  equal  to 

a-\-  b  +  c 


and  the  original  equations  become 

a  b  c    ^, 

Putting  ^(a  +  &  +  c)  in  plac*""  c    k  in  the  results  of  Ex.  1,  we 

have 

a:  =  i(a  +  6  +  c) (-  a  +  6  +  c), 

2^=  i(a  +  6  +  c)(a-6  +  c), 
0=  i(a  +  6  +  c)(a  + 6-c).    • 


I 


EQUATIONS   WITH   FRACTIONS.  227 

Ex.  3.   Solve  the  simultaneous  equations 

y-\-Sz_z  +  x_Sx  +  5y_  ^ 
4       ~      5     ~        2 

By  Theorem  I.,  page  208,  we  have 

6y  +  l^z  —  16z  —  lBx  —  Sx  —  by_  ^ 
20-75-2  ' 

.-.  -18x  =  -57,  x  =  f|; 
y  +  Zz-Sz-Sx  +  Bz  +  6y_^ 
4-15  +  2 

.-.  6y  =  -9,  y  =  -i; 
6y  +  15g  +  3g  +  3a;-3a;-6y_  ^ 
20  +  15-2 

.-.  180  =  33,  z  =  ^. 


EXAMPLES  LI. 

Solve  the  following  simultaneous  equations,  obtaining  one  set  of 
values  of  the  unknown  quantities  in  each  case. 

-  1      1     X     ?/      2       2 

^     a     h    a     h     a^     y^ 


2. 


X y_-^,  __y a;    _  1 

a  +  b     a  —  b     ab    a  -{•  b     a—  b     ab 


3. 

a     b 

:1,    f  +  ^  = 

6     a 

1. 

4. 

X 

x-\-\ 

-.-It' 

1 
x  +  l 

6. 

X 

■-^-1, 

.:  -\-y  =  a 

+  &. 

x-a     y 

6.  ?  +  I^  =  l,  ^  +  i^  =  l. 

a     b     c  a'     b'     c' 

7.  _  +  -  =  c,  —  +  -  =  c'- 
X     y  x      y 


228  EQUATIONS   WITH  FRACTIONS. 


X  Sy  xy 

X     y       '  y     z       '   g^ic 

10.  1  +  1  =  7,   1  +  1  =  9,   1  +  1  =  12. 

y     z  z     X  X     y 

11.  '^  =  y  =  ^  =  (x  +  y  +  zy- 
a     0     c 

12    3y-^-ig-3a;_a;+y_2 
2  3  5' 

a  6  c 

J,  ^    cy  +  &^  _  az  +  ca;  _  6a;  +  ay  _  i.  * 
Z  m  n 

15.   Pind  two  sets  of  values  for  aj,  ?/,  0,  that  satisfy  the  equations 
X  _y  _z _        1 


a     &      c     x  +  y  +  z 

16.  Find  three  sets  of  values  for  x,  y,  z,  that  satisfy  the  equa- 
tions 

^  =  y=^  =  xyz. 
a     b     G 

17.  Find  a  value  of  c  that  will  make  the  equations 

2a;2-a:  +  c=:0,  4a;2+4a:  +  l  =  0 
have  a  common  root,  and  then  determine  the  common  root. 

18.  Determine  two  values  of  a,  either  of  which  will  make  the 
equations 

a;-2  +  2 a;  -  3=  0,  aa;2  -  11 X  +  10  =  0 

have  a  common  root,  and  find  the  common  root  in  each  case. 
*  See  Ex.  3,  of  Art.  170. 


MISCELLANEOUS    EXAMPLES   III.  229 

MISCELLANEOUS  EXAMPLES   III. 

A.  1.  What  must  be  added  to  x^  —  3  x^y  +  3  xy^  that  the  sum 
may  be 

Sx^y-Sxy^  +  y^? 

2.  Show  that 

x(y  -i-  zy-^  y  (z  +  xy  +  z  (x  +  yy  -  4  xyz  =  (y+z)(z  +  x)  (x  +  y). 

3.  Find  the  factors  of  (l)  x^  -  y^,  (ii.)  a^ -b^ -\- 2a -2b,  and 

(iii.)  x^-Sx-2S. 

4.  Find  the  H.  C.  F.  of  x*  +  9a;  -  20  and  5a;*  +  9x^  -  64. 
Simplify  ^^  +  ^^-^Q    . 

6.  Solve 

(ii.)  Sx-4y-\-2z=5x-6y-2=7x-\-2y-\-i. 

(iii.)  -JL-4.     ^     =^L±A. 
x  +  b     x  +  a         X 

6.  If  A  were  to  give  $  10  to  B,  he  would  then  have  three  times 
as  much  as  B  ;  but  if  B  were  to  give  $  5  to  A,  A  would  have  four 
times  as  much  as  B.     How  much  has  each  ? 

B.  1.   Find  the  value  of 

a  (a  +  6)  (a  +  6+ c)  —  a  (a  -  6)  (a  —  6  —  c), 
when  a  =  5,  6  =  3,  and  c  =  —  6  ;  also  when 

a  =  —  3,  6  =  —  2,  and  c  =  4. 

2.   Show  that  a^  +  3(a  - 2 6)2  =  S(a-by  +  (a-S by,  and  that 

8.  Find  the  factors  of  (i.)  x^  -4:X^  +  ix,  (ii.)  2  a;2  _  6  -^  +  2,  and 
(iii.)  asS  -  a;2  +  a;  -  1. 


230  MISCELLANEOUS   EXAMPLES   III. 


lift 
4.    Simplify  -—  +  ---^--^-^,  and  show  that 

\        y-x  /\        y -X  J     \ y -X  J 


y -X  J \        y 
6.   Solve 


X 


=  1 


x—1     X — 2     X— 3  x.y_a 

ha     b  _ 

6.  Find  the  fraction  which  becomes  ^  when  2  is  added  to  its 
numerator,  and  ^  when  2  is  taken  from  its  denominator. 

C.   1.   Prove  that 

(a;-  1)2(2/2  +  i)_(a;2  +  l)(2/  _  1)2  =  2(x  -  y){xy  -  1), 
and  that 

{X  +  y)  (a;  +  2)  -  x2  =  (y  +  z)  (2/  +  x)  -  2/2  =  (0  +  x)  (^  +  y)  -  ^2. 

2.  Prove  that  the  product  of  any  two  numbers  is  equal  to  a 
quarter  of  the  difference  of  the  square  of  their  sum  and  the  square 
of  their  difference. 

3.  Find  the  factors  of  (i.)  1  +  18 cc  -  63 a;2,  (ii.)  3 a:^?/ -  24  y*, 
and   (iii.)  (x^  ^  2x)^ -{Bx^ +  iy. 

4.  Simplify  ^^-^Q^  +  ^\ 

and  1  ^-4  '  ^-^ 


(x-3)(a;-2)      (a;-l)(x-3)      (x-l)(a-2) 

6.  Solve 

^' \  x  —  S     x  —  l  I  x  +  1     x  —  S_f^ 
00-^  i~'^"6  8~-^' 

(ii.)  ax  +  by  =  c,  a2ic  +  b^y  =  c^. 

6.  A  and  B  have  $  100  between  them  ;  but  if  A  were  to  lose  one- 
half  of  his  money,  and  B  were  to  lose  one-third  of  his,  they  would 
then  have  only  $  55  between  them.    How  much  has  each  ? 


MISCELLANEOUS   EXAMPLES  III.  231 


D.   1.  Find  the  value  of    Va^  -{- b^  +  c^  -{a  -  b  -  c)2,  when 
a  =  3,  6  =  —  3,  and  c  =  4. 

2.  Show  that 
(x-5)(x-4)-3(sc-2)(a;-l)+3(x+l)(a;+2)-(x+4)(a;+5)=0, 

and  that 

1  +  n  +  Jn(n  +  1)+  in(«  +  l)(n  +  2)  =  Kw  +  l)(n  +  2)(n  +  3). 

3.  Find  the  factors  of  (i.)  x^ + x^y  -  6  xy%  (ii. )  x^ + a5c2  -  a^a;  -  a', 
and  (iii.)  xV  _  a;2  -  y2  _|.  i. 

4.  Simplify 

a  ^  ^^  ~  ^  I     4  ax     ,  g  +  a; 
a  +  X     a^  _  a;2     ^  -  x 

^    ^  x-4     x-5     x-6 
6.  Solve 

(i.)  3x+i_2(6-x)  =  ^^^-^^. 


(ii.)  3y  +  ?  +  6  =  0,    y-]--  =  S. 

X  X 

6.  A  hare  is  pursued  by  a  greyhound,  and  is  60  of  her  own  leaps 
before  him.  The  hare  takes  3  leaps  in  the  time  that  the  greyhound 
takes  2  ;  but  the  greyhound  goes  as  far  in  3  leaps  as  the  hare  does 
in  7.     In  how  many  leaps  will  the  greyhound  catch  the  hare  ? 

E.  1.   Show  that 

(m  +  n)(^m  +  n  —  1)  =  w(m  —  1)+  2mn  +  n(n  —  1), 
and  that 
(m  +  n)  (m  +  n  —  1)  (m  +  n  —  2)  =  m(m  —  1)  (m  —  2) 

+  3wi(m-  l)n  +  3mn{n  -  1)+  n{n  -  l)(n-2). 

2.  Divide  x*  —  y*  by  x  —  y,  and  from  the  result  icrite  down  the 
quotient  when  (x  +  yy  —  IQz^  is  divided  hy  x  +  y  ~2z. 


232  MISCELLANEOUS   EXAMPLES   in. 

3.  Find  the  factors  of 

5  x2  +  24  oj  -  6,  and  of  a^  -  2  ahc  -  ah"^  -  m'^. 

4.  Find  the  L.  C.  M.  of  8  a^  _  is  aV^,  8  a^  +  8  a^d  -  6a62,  and 
4  a2  -  8  a6  +  3  6-2. 

6.    Show  that 

(i.)l4--^+  ^^  -  ^' 


X  —  a     {x  —  a){x  —  h)      {x  —  a){x  —  h) 


(ii.)  !+_«-+  "^ 


X  —  a     {x  —  a){x  —  h)      {x  —  a){x  —  6)  (x  —  c) 

^ ^ 

{x  —  a)(x  —  b)  (x  —  c) 

6.  A  debtor  is  just  able  to  pay  his  creditors  25  cents  on  the 
dollar ;  but  if  his  assets  had  been  five  times  as  great,  and  his  debts 
two-thirds  of  what  they  really  were,  he  would  have  had  a  balance 
in  his  favor  of  $  700.     How  much  did  he  owe  ? 

F.    1.   Find  the  value  of 

{a~{b-  c)}2  +  {6  -(c  -  a)}2  +  {c-(a-  b)f 
when  a  =  —  1,    6  =  —  3,   and  c  =  —  5. 

2.  Divide  x^  -  2  a*x*  +  a^  by  x^  +  ax^  +  a^x  +  a^. 

3.  Find  the  L.  C.  M.  of  x^  -  1,  (x  -  1)2,  (x  +  1)2,  x»  -  1,  and 
xS+1. 

4.  Add  together  »    >    and — :    and  show 

2x  +  2/     2x-y  y2_4a;2' 

that 

(x  -a)(y-  a)      (x-b)(y-b)      (x  -  c) (y  -  c)  ^ 
(a  -  6)(a  -  c)  "^  (6  -  c)(6  -  a)  "^  (c  -  a)(c  -  6) 

5.  Solve  (i.)  -^—+     ^  ^ 


x  +  a     x  +  6     x  +  a  +  6 
(ii.)  5x  +  22/  +  3^  =  13j 
Sx  +  7y-z=   2  ^« 
X  — 2y  +  2=   sJ 


MISCELLANEOUS    EXAMPLES   III.  233 

6.  When  the  arable  land  of  a  farm  was  let  at  $  7.50,  and  the 
pasture  at  $10  an  acre,  the  total  rent  of  a  farm  was  $2750. 
When  the  rent  of  the  pasture  was  reduced  by  $  1.25  an  acre,  and 
that  of  the  arable  land  by  $2.50  an  acre,  the  whole  rent  was 
$  1937.50.    What  was  the  total  acreage  of  the  farm  ? 

G.   1.    Solve  the  simultaneous  equations 

X  —  a  .      y  —  h     _ ^ 

c         a  +  c  —  b       ' 

x-\-  a  —  c  .  y  —  a_ 


a  +  c        a  —  b     a  +  c 

2.  Solve  the  simultaneous  equations 

yJ:S-a=^±^-b='^^±y-c  =  x  +  y  +  z. 
a  b  c 

3.  Solve  the  simultaneous  equations 

^/y  +  c/g _ c/z  +  a/x _ a/x  +  b/y _a  .  b  .  c     ^ 
a  b  c  X     y     z       ' 

4.  A  man  rides  one-half  of  the  distance  from  A  to  B  at  the  rate 
of  a  miles  an  hour,  and  the  other  half  at  the  rate  of  b  miles  an  hour. 
If  he  had  travelled  at  the  uniform  rate  of  c  miles  an  hour,  he  would 
have  ridden  the  whole  distance  in  the  same  time.    Prove  that 

2  =  1+1. 
cab 


234  QUADRATIC   EQUATIONS. 


CHAPTER  XVI. 
Quadratic  Equations. 

175.  Let  it  be  required  to  state  the  following  problem 
in  the  form  of  an  equation. 

If  the  speed  of  a  certain  train  were  increased  by  4 
miles  per  hour,  the  time  it  now  requires  to  make  a  trip 
of  120  miles  would  be  decreased  by  one  hour.  What  is 
the  train's  speed  in  miles  per  hour  ? 

Let  X  =  the  speed  of  the  train  in  miles  per  hour.    Then 
120/ic  =  the  number  of  hours  the  train  now  requires 

for  the  trip  of  120  miles ; 

120/(a;-|-4)  =  the  number  of  hours   it  would  require 

for  the  same  trip,  were  its  speed  increased  as  indicated, 

and  by  the  conditions  of  the  problem  this  is  one  hour 

less  than  120/a;  hours ; 

120       120 


X-i-4: 


■M. 


Multiplying  both  sides  of  this  equation  byic(a;+4), 
we  have 

120  a;  +  480  =  120^  -{-3^  +  4:X, 

whence  ic^  +  4  ic  =  480, 

The  problem  of  finding  the  speed  of  the  train  is  thus 
reduced  to  that  of  solving  an  equation  which  contains 
both  the  first  and  second  powers  of  its  unknown  quantity. 


QUADRATIC   EQUATIONS.  235 

Some  equations  of  this  class  were  considered  in  Arts. 
128,  129.  We  now  proceed  to  a  more  complete  analysis 
of  them. 

176.  A  quadratic  equation  is  an  equation  which  contains 
the  unknown  quantity  to  the  second,  but  to  no  higher 
power.'  Thus  x^  =  4:,  3a?'+4a;=7,  and  ax^  +  5rc  +  c  =  0 
are  quadratic  equations. 

When  all  the  terms  are  transposed  to  one  side,  a  quad- 
ratic equation  must  be  of  the  form 

ax^  -{-  hx  -\-  c  =  0, 

where  a,'  b,  c  are  supposed  to  represent  known  quantities. 
It  was  shown  in  Art.  122  how  the  factors  of  the 
expression  on  the  left  side  of  this  equation  can  be  found 
by  chayiging  the  expression  into  an  equivalent  one  which  is 
the  difference  of  two  squares. 

We  first  apply  the  method  to  some  examples ;  but  the  student 
will  profit  by  reading  Art.  122  again  before  considering  them. 

Ex.  1.     Solve  x^  +  12  X  +  35  =  0. 

Since  x^  +  12  a;  is  made  a  perfect  square  by  the  addition  of  the 
square  of  half  the  coefficient  of  x,  that  is  the  square  of  6,  we  add 
and  subtract  62.     The  equation  then  becomes 

x2  +  12  X  +  62  -  62  +  35  =  0, 

that  is  (x  +  6)2  -  1  =  0  ; 

...  ((x+6)+l}{(x4-6)-l}  =  0, 

that  is  (x  +  7)  (x  4-  6)  =  0. 

Hence  x  +  7  =  0,  or  x  +  5  =  0. 

Thus  X  =  —  7,  or  X  =  —  5. 

Ex.  2.     Solve  3  x2  =  10  X  -  3. 

By  transposition  we  have 

3x2-10x  +  3  =  0. 


236  QUADRATIC   EQUATIONS. 

We  must  first  divide  by  3  in  order  to  make  the  coefficient  of  y? 
unity  ;  the  equation  then  becomes 

Half  the  coefficient  of  a;  is  —  f ,  and  therefore  x^  —  J^O-  x  will  be 
made  a  perfect  square  by  the  addition  of  (—  |)2  ;  and  adding  and 
subtracting  (—  1)^,  that  is  ^-^  the  equation  becomes 

...   (X- 1)2 -1^  =  0, 

that  is  (x_  1)2 -(4)2^0; 

...  (a;_5  +  |||^_5_||^0; 

...   (x-i)(x-3)=0.  .       ., 

Hence  x  —  |  =  0,  or  oj  —  3  =  0. 

Thus  ic  =  ^,  or  a;  =  3. 

Note.  — In  many  cases  the  factors  can  be  written  down  at  once, 
as  in  Arts.  117  and  118,  without  completing  the  square.  The 
student  should  always  see  if  this  can  be  done :  he  will  thus  save 
himself  a  great  deal  of  unnecessary  work. 

Ex.  3.     Solve  4  ic  -  x2  =  2. 

By  transposition,  we  have 

4x-ic2_2  =  0. 
Change  all  the  signs  so  as  to  make  the  coefficient  of  x'^  unity ; 
then  we  have 

cc2-  4ic  +  2  =  0. 

Add  and  subtract  4,  the  square  of  half  the  coefficient  of  x.     Then 
x2-4x  +  4-4  +  2=0; 
.-.  (x- 2)2 -2  =  0; 
that  is  (ic  -  2)2  -  (  ^2)2  =  0  ; 

...  (x-2+V2)(x-2-V2)  =  0. 
Hence  x  -  2  +  ^2  =  0,  or  x  -  2  -  ^2  =  0. 
Thus  X  =  2  -  v/2,  or  X  =  2  +  V2. 


QUADRATIC  EQUATIONS.                         287 

EXAMPLES  LIT. 
Solve  the  following  equations : 

1.  a;^-4x  =  45.  22.  17aj2  +  8  =  70a;. 

2.  x^-6x  =  91.  23.  51x2  4-19z  =  10. 

3.  (a;-3)2  =  x  +  3.  24.  110x2  -  21a; +  1  =  0. 

4.  (x-4y  =  x-2.  25.  21x2 -13 a;  =  20. 
6.    (x-l)(x-2)  =20.  26.  21x2 +  23 x  =  20. 

6.  (x  +  l)(x  +  3)  =  2(x  +  3).  27.  6x2  +  6  =  13x. 

7.  4x  +  3  =  x(x  +  2).  28.  6x2  =  5x+l. 

.8.  4x-3  =  x(2-x).  29.  9x2  -  63x  +  68  =  0. 

9.    (x+l)8=(x-l)8  +  26.  30.  16x2  +  3=  16x. 

10.  (x-l)8  =  (x  +  l)8-56.  31.  29x2 -41x- 138  =  0. 

11.  9x2  +  6x  +  l=0.  32.  29x2 +  11  x  =  138. 

12.  9x2  +  16  =  24x.  33.  x2  -  16  =  216  -  lOx. 

13.  x2+150  =  53x.  34.  (x  +  2)2  =  4(x  -  1)2. 

14.  x2  =  2  X  +  99.  86.  (x  +  6)2  =  16(x  -  6)2. 
16.   3x2  +  3  =  10x.  86.  (x  + 8)2  =  9x2. 

16.  3x2  +  11 X  =  20.  87.  (x  -  7)2  =  49(x  +  2)2. 

17.  4x2+21x  =  18.  88.  x2  -  3ax  +  2a2  =  0. 

18.  6x2  +  65x  =  50.  89.  x2  +  3a2  =  4ax. 

19.  24x2  =  30x+75.  40.  x2  +  2a6  =  62  +  2ax. 

20.  x2  -  200  =  35x.  41.  4 x2  +  4  ax  =  62  _  a^ 

21.  19x2-39x  +  2  =  0.            „  42.  x2  +  2ax  =  62  +  2a6. 

177.   To  solve  the  equation 

ax^  -\-bx-{-  c  =  0. 

Divide  by  a,  the  coefficient  of  ar';  the  equation  then 
becomes 


238  QUADRATIC   EQUATIONS. 

Now  add  and  subtract  the  square  of  half  the  coefficient 

of  X,  that  is  the  square  of  -  —    Then  we  have 

Z  a 

a        \2aJ      \2aJ      a 
The  first  three  terms  are  now  a  perfect  square,  namely, 

Hence  we  have  (x  -\ )  —  ( — -  —  -  )=  0, 

V       2aJ      \4:a^     a) 

Hence  [Art.  115] 

We  therefore  have 

2a  2a  2a 

b       -^{W-^ac)  _~h-^{W-4.a6) 

or  flJ  =  —  pr pr  — ^ • 

2a  2a  Za - 

It  follows  from  the  above  that  a  quadratic  equation  has 
two  and  only  two  roots. 


QUADRATIC   EQUATIONS.  239 

178.  Instead  of  working  out  every  example  from  the 
beginning,  we  may  use  the  general  formula  found  in 
Art.  177,  and  substitute  for  a,  6,  and  c  their  values  in  the 
particular  equation  we  are  considering. 

Thus,  the  roots  of  the  equation  ax^  +  bx  +  c  =  0  being 

2a  2a 

we  find  the  roots  of  3  ic^  -  10  x  +  3  =  0  by  putting  3  for  o,  -  10 
for  6,  and  3  for  c  in  the  above  formula. 
Hence  the  roots  of  3  aj^  —  10  x  +  3  =  0  are 
-10      V(100-36) 
6  6   ' 

that  is  \°  ±  I ;  and  hence  the  roots  are  3  and  \. 

179.  General  Properties.  In  discussing  the  quadratic 
equation  aa^  -\-  bx  -\-  c  =  0,  it  is  assumed,  unless  a  contra- 
dictory hypothesis  be  made,  that  the  coefficients  a,  b,  c 
are  finite  and  different  from  zero ;  and  the  general  prop- 
erties of  this  equation  are  enunciated  under  this  limi- 
tation. 

In  Art.  177  we  found  that  the  equation  ax^  -f  5x  -f  c  =  0 
had  two  roots,  namely, 

b       v(&^-4ac)  b       V(6^-4ac) 

2a^  2a  2a  2a 

(i.)  If  b^  —  Aac  be  positive,  \^(b^  —  4ac)  is  the  square 
root  of  a  positive  quantity  and  has  a  numerical  value, 
either  rational  in  the  form  of  an  exact  square  root,  or 
irrational  in  the  form  of  an  indicated  square  root.  The 
quantity  -y/(6*  — 4ac)  is  here  said  to  be  real.  The  roots 
of  the  quadratic  equation  are  then  also  real,  and  they 
are  obviously  different  from  one  another.     Also,  unless 


240  QUADRATIC   EQUATIONS. 

6^  —  4ac  be  positive,  V(^^  —  4ac)  has  no  numerical  value 
and  is  not  real.     Hence : 

In  order  that  the  roots  of  ax^  +  6a;  +  c  =0  may  he  real,  it 
is  necessary  and  sufficient  that  6^  >  4  ac. 

From  the  formula  of  Art.  177  it  is  clear  that  the  roots 
of  the  equation  aa?  -\- hx -\-  c  =  0  are  irrational  unless 
6^  —  4  ac  is  a  perfect  square. 

It  is  also  clear  that  if  either  of  the  roots  of  a  quad- 
ratic equation  be  irrational,  they  are  both  irrational. 

Let  the  student  here  call  to  mind  the  meaning  of  the 
expression  irrational  quantity.  It  is  a  quantity  whose 
exact  value  cannot  be  expressed  either  as  an  integer,  or 
as  a  fraction  whose  numerator  and  denominator  are 
integers.  In  algebra  it  is  usually  expressed  symbolically, 
as  in  the  case  under  consideration  by  means  of  the  sign 
V-     [See  Art.  20.] 

(ii.)  If  6^  —  4ac  =  0,  both  roots  reduce  to  —  h/2a,  and 
are  thus  equal  to  one  another.  In  this  case  we  do  not 
say  that  the  equation  has  only  one  root,  but  that  it  has 
two  equal  roots.  Also,  the  roots  are  unequal  unless 
6^  — 4ac  =  0.     Hence: 

In  order  that  the  roots  of  ao?  -\-hx-{-c  =  0  may  he  equal 
to  one  another,  it  is  necessary  and  sufficient  that  6^  —  4  ac  =  0. 

When  6^  —  4  ac  =  0,  the  expression  ax^  -[-hx-\-c  is  a 
perfect  square  so  far  as  x  is  concerned;  for  ax^-\-hx-{-  c, 

that  is,  a(a^  +  -x-{--\  will  then  be  a(x-\ ),   since 

^,       ^       \        a        aj  \       2aJ 

4a^     a 

(iii.)  When  &^  — 4ac  is  negative,  y(6^  — 4ac)  is  the 
indicated  square  root  of  a  negative  quantity,  and  has  no 


<^ 


QUADRATIC   EQUATIONS.  241 

numerical  value.  It  is  therefore  excluded  from  the 
category  of  so-called  real  quantities,  and  is  called  an 
imaginary  quantity.    [Art.  228.]     Hence  : 

In  order  that  the  roots  of  ax^  +  6a;  +  c  =  0  may  he  imagi- 
nary, it  is  necessary  and  sufficient  that  h^<,^ac.  ' 

It  should  be  noticed  that  if  either  of  the  roots  of  a 
.quadratic  equation  be  imaginary,  they  are  both  imaginary. 

180.  The  above  criteria  of  the  character  of  the  roots 
of  a  quadratic  equation  are  more  concisely  stated  thus : 

(1)  If  6^  — 4ac  be  positive,  the  roots  are  real  and 
unequal. 

(2)  If  6^  —  4ac  =  0,  the  roots  are  equal. 

(3)  If  6^  — 4ac  be  negative,  the  roots  are  imaginary* 
and  unequal. 

(4)  If  6^  — 4ac  be  a  perfect  square,  the  roots  are 
rational  and  unequal. 

(5)  If  6*  —  4  cic  be  not  a  perfect  square,  the  roots  are 
irrational  and  unequal. 


181.  By  multiplying  both  sides  of  an  equation  by  the 
same  quantity,  we  do  not  destroy  the  equality,  and 
therefore  the  new  equation  is  satisfied  by  all  the  values 
which  satisfy  the  original  equation;  if,  however,  we 
multiply  by  any  integral  expression  which  contains  the 
unknown  quantity,  the  new  equation  will  be  satisfied  not 
only  by  the  values  which  satisfy  the  original  equation, 

*  Complex  quantity  is  a  more  general  and  more  accurate  name  for 
expressions  of  the  form  in-\-  V^n.    The^term  imaginary  is  then  ap- 
plied only  to  expressions  of  the  form  V—  n.    [See  Chapter  XXII.] 
Q 


242  QUADRATIC   EQUATIONS. 

but  also  by  any  value  which,  makes  the  expression  by 
which  we  multiply  vanish. 

For  example,  the  equation  x^  =  9  is  satisfied  by  the  values 
X  =  3,  or  x  =  —  3.  If  we  multiply  both  sides  of  the  equation  by 
aj  —  2,  we  have  the  new  equation 

and  this  new  equation  is  satisfied  not  only  by  x  =  3,  and  by 
X  =  —  3,  but  also  by  X  =  2, 

Thus  additional  roots  are  introduced  whenever  both 
sides  of  an  integral  equation  are  multiplied  by  any  inte- 
gral expression  which  contains  the  unknown  quantity. 

When  an  equation  contains  fractions  in  whose  de- 
nominators the  unknown  quantity  occurs,  the  equation 
may,  however,  be  multiplied  by  the  L.  C.  M.  of  the 
denominators  without  introducing  any  additional  roots ; 
for  we  cannot  divide  both  sides  of  the  resulting  equation 
by  any  one  of  the  factors  of  the  L.  C.  M.  without  re-intro- 
ducing fractions. 

The  student  must,  however,  be  careful  to  multiply  by 
the  lowest  common  multiple  of  the  denominators,  for 
otherwise  the  resulting  equations  will  have  roots  which 
are  not  roots  of  the  original  equation. 

We  may,  for  example,  multiply  both  sides  of  the  equation 

2x  10     ^     7 

x-1      x2-l     x-t-1 

by  x2  —  1,  the  L.  C.  M.  of  the  denominators  of  the  fractions ;  we 
thus  obtain 

2x(x  +  l)-10  =  7(x-l), 

and  this  new  equation  is  not  satisfied  by  either  of  the  values 
obtained  by  equating  to  zero  the  factor  by  which  we  have  multi- 


QUADRATIC   EQUATIONS.  243 

plied,  and  hence  no  additional  roots  have  been  introduced  by  the 
multiplication. 

If,  however,  we  multiply  the  original  equation  by  (x-l)(x2  — 1), 
we  shall  have 

2x(x+l)(x-l)-10(x-l)=7(x-l)2,   . 

and  one  of  the  roots  of  this  equation,  namely  x  =  1,  is  not  a  root  of 
the  original  equation. 

Note.  —  The  student  should  always  remember  that  when  both 
sides  of  an  equation  are  divided  by  any  factor  ichich  contains  the 
unknown  quantity^  the  resulting  equation  will  not  give  all  the 
roots  of  the  original  equation  ;  to  find  the  remaining  roots  we  must 
equate  to  zero  the  factor  by  which  we  have  divided. 

For  example,  if  we  divide  both  sides  of  the  equation 

(x2  -  4)(x  +  2)  =  (x2  -  4)(2x  +  1) 
by  x2  —  4,  we  obtain  the  equation 

x  +  2  =  2x-l, 

from  which  we  obtain  x  =  3.  But  x  =  3  is  not  the  only  solution  of 
the  original  equation  ;  to  obtain  the  other  solutions  we  must  equate 
x2  _  4  to  zero. 


EXAMPLES  LIII. 

Solve  the  following  equations  : 

X.  ^.H 

g    x+lx  +  2_29 
•   X  +  2     X  +  1      10 

-  -i=f- 

7       1            2     _3 

*  x-2     x  +  2     5 

2     a;      3 

8        '       1      ^     -^ 

x+4'4-x     3 

.     x+5        4     . 
'       4         X  +  & 

_3, 
2* 

9.      \4-     ^=3. 

x-2     x+1 

6.       \+==-V 

X-  1             X 

13 
6* 

10       5             10      _      2 

'x  +  1     x  +  10     3x-3 

244  QUADRATIC   EQUATIONS. 


11. 

i+     1     +      1     =0. 

2      3  +  X     2  +  3x 

17. 

2x-l      13_3x  +  5 
2x+l      11     3x-5 

12. 

1+     ^     +      ^      -0. 

3^j»  +  3     x  +  n 

18. 

2x-2     3-3x_       5 

2x-3     3x-2     8x-12 

13. 

.^x-1      2^+1  =  5. 
x+l        x-1 

19. 

3x            4             4     _o^ 
x-2     x+3     2-x 

14. 

3            4     _    15 

x-1     x-3     x+3 

20. 

5x           6            19    _Q 
x-3     x+2     3-x 

16. 

4x  +  3_6x-13     7x-46 
9              18           5x  +  3 

21. 

x-lx+l_    5x 

x+l        x-1        X2-1 

16. 

2x-3_x-5     5X-16 
4            12     '     x-1 

22. 

1          +          1           =    ^• 

x2-3x     x2  +  4x     2x2 

A 

23.        ^ 

x2-3x     9 

J 

■X2" 

13 
16  X 

«1         1       I 

5 

J_ 

7X                   _Q 

/^-  Sx-eU 

- 

x^' 

72(x  +  2) 

25. 

1             17         1 

27. 

1+      1     =     ^     +      ^     • 

X     x+4     x+l      x+2 

x2-l      1-x     8     x+l 

26. 

1             3^1 

28. 

X        x+l_x-2      x-1 
x+l     x+2     X— 1        X 

a.2__4     2-x         '3x  +  6 

09       ^         a:- 

3 

_{_ 

X         x+3_2 

*  x-3         X 

X 

+  3x3 

80.^  +  ^  +  ^- 

X+l       X  — 

6. 
4 

X- 

+  4     x-5 

X- 

+  2     x-3 

31.^-3     X- 

J 

_x 

-6     x-7 

X  —  4x  —  5x  —  7x  —  8 

32.  x2  +  2  a2  :=  3  ax.  35.    a(x2  +  1)  =  x(a2  +  i). 

33.  9 x2  -  6  ax  =  a2  -  &2.  36.    a;2  -  2(a  -  6)x  +  &2  =  2  ab. 

34.  a(x2  -  1)  +  x(a2  -  1)  =  0.        37.   x2  +  2(6  -  c)x  +  c2  =  2  be. 

38.   (b  -  c)x2  +  (c  -  a)x  +  (a  -  6)  =  0. 


QUADRATIC  EQUATIONS.  245 

39.  (a  +  6)x2  +  c«-a-6-c=0.      41.    fecx^  +  (62  +  c2)x  +  6c  =  0. 

40.  a6a;2-(a2  +  62)a;  +  a6  =  0.      42.    (a'^  -  b^)  {x^  -  1)  =  4:  abx. 

43.   (62  -  o2)(a:2  +  1)  =  2(a2  +  62)a;. 


44.   x  +  -  =  a-\-^ 
a  X 


49.   tl^ +  -=-  +  , 
a     X     a     0 


45.   _^  +  _^  =  4.  50.   1_-  =  1  +  1  +  1. 

a+x^a-x  x+a+b     x     a     b 

46    _l_  +  _i_  =  l  +  l.  61.   __L_  =  1-1_1. 

■   x-a     x-b     a     b  x-a-b     x     a     b 

47.  _«L.  +  _^  =  «  +  ^.  52.   J-4-l-=-i-+-^ 
x-a     x-b      b     a  x+a    x+6    c+a    c+6 

48.  ^i-  +  ^^  =  a  +  6.  53.  -£-+-^=-^+-^. 
x-6     X  — a  x+a    ic+6    c-\-a    c-{-u 

g.    x  +  q  ■  x  +  b  _x  —  a  .  x-b 
'  x  —  a     x  —  b     x  +  a     x  +  b 

65.   -J^  +  -^=     ^-^     +     ^-^^    ♦ 
x + a     x+6     x+a-c     x+6+c 


182.  Irrational  Equations.  An  irrational  equation  is 
one  in  which,  square  or  other  roots  of  expressions' con- 
taining the  unknown  quantity  occur. 

In  order  to  rationalize  an  equation  it  is  first  written 
with  one  of  the  irrational  terms  standing  by  itself  on 
one  side  of  the  sign  of  equality :  both  sides  are  then  raised 
to  the  lowest  power  necessary  to  rationalize  the  isolated 
term ;  and  the  process  is  repeated  as  often  as  may  be 
necessary.  J 

Ex.  1.   Solve  the  equation  v^x2  _  9)+x  =  9. 
By  transposition,  we  have 

y'(x2-9)  =  9-a;. 


246  QUADRATIC   EQUATIONS. 

Square  both  sides  ;  then 

x2  -  9  =  (9  -  a;)2 ; 
.-.  18x  =  90;   .-.  x  =  6. 

Ex.  2.    Solve  the  equation 

V(2x+8)-2  V(a^  +  6)=2. 
Squaring  both  sides,  we  have  \ 

2x  +  8  +  4(a:  +  5)-4v(2a:+8)V(a;  +  6)  =  4, 
that  is  6a;  +  24-4v(2x  +  8)V(a;  +  5)  =0; 

.-.  3a:  +  12  =  2V(2a;  +  8)V(a;+5). 
Square  both  sides  of  the  last  equation,  and  we  have 
9x2  +  72x  +  144  =  4(2x  +  8)(x  +  6); 
.-.  x2-16  =  0. 
Hence  x  =  4,  or  x  =  —4. 

Note.  — If  we  put  4  for  x  in  the  original  equation,  we  should 
have  ^16  _  2^9  =  2,  that  is  4  -  6  =  2, 

which  is  not  true. 

Again,  if  we  put  —  4  for  x,  we  should  have 
V0-2V1=2, 
which  is  not  true. 

Thus  neither  of  the  values  we  found  for  x  appears  to  satisfy  the 
equation. 

The  failure  is  due  to  our  having  supposed  that  V(2x  +  8)  and 
■y/{x  +  5)  were  necessarily  positive,  whereas  every  algebraical 
quantity  has  two  square  roots,  one  positive  and  the  other  negative, 
and  the  square-root  symbol  does  not  necessarily  represent  one 
only  to  the  exclusion  of  the  other. 

Bearing  this  in  mind  we  shall  find  that  x  =  4  does  satisfy  the 
equation,  if  the  ambiguous  signs  be  inserted  ;  for  the  condition  is 

±y/16-  2(±V9)  =  2,  that  is  ±4 -(±6)=  2, 

which  is  true  when  the  lower  signs  are  taken. 


QUADRATIC   EQUATIONS.  247 

If  an  irrational  equation  is  to  be  solved  on  the  supposition  that 
the  radicals  it  contains  are  to  be  necessarily  positive  {or  necessarily 
negative),  the  equation  will  be  accompanied  by  a  definite  statement 
to  that  effect. 

Ex.  3.     Solve  the  equation 

V(2x  +  7)  +^(3x  -  18)  +y/(7x  +  1)  =  0. 

Since  the  sum  of  three  positive  numbers  cannot  be  zero,  this  is 
not  a  possible  equation  unless  one  or  more  of  the  square  roots 
involved  in  it  be  negative. 

By  transposition,  we  hkve 

V(2  X  +  7)  -Jv(3  X  -  18)  =  -  V(7  a;  +  1). 
Square  both  members  I  then  I 

2x  +  7  +  2V(2a;  +  7)  V(3a:  -  181+  3a;  -  18  =  7x  +  1. 
By  transposition  and  division  by  2,  wb  have 

V(2a:  +  7)V(3x-18)=a;  +  6. 
Square  both  members  ;  then 

(2a;  +  7)(3x-18)=(a;  +  6)2; 
.-.  6a;2-27a;-162  =  0. 
Hence  a;  =  9,  or  a;  =  —  y-. 

By  applying  the  test  for  a  solution  [Art.  91]  it  will  be  found 
that  X  =  9  is  a  solution  when  ^^(7  x  +  1)  is  taken  negatively,  and 
that  X  =  —  18/5  is  a  solution  when  y/{Sx  —  18)  is  taken  negatively, 
the  two  remaining  square  roots  being  taken  positively  in  each  case. 
The  process  of  rationalizing  equations  introduces  factors  that 
contain  the  unknown  quantity.  Hence,  in  solving  irrational  equa- 
tions it  is  important  to  apply  the  test  for  a  solution.    [See  Art.  181.] 

Ex.  4.     Solve  the  equation 

-\-^(x^-9)-(x-9)=0, 
wherein  +  y/(x'^  —  9)  indicates  that  only  the  positive  square  root  is 
to  be  taken. 

Multiplying  by  +y/(x'^  —  9)-{-(x —  9),  we  have 
a;2_9_x2-f  18x-  81  =0; 
whence  18  x  =  90,  and  x  =  5. 


248  QUADRATIC   EQUATIONS* 

But  5  is  not  a  root  of 

+  V(^2_9)_(a._9)^0, 
and  it  is  a  root  of  the  equation 

by  whose  left  member  the  original  equation  was  multiplied.  It 
may  be  shown  that  there  is  no  finite  number  that  will  satisfy  the 
equation 

+  V(^"^-9)-(a^-9)=0. 

EXAMPLES  LIV. 

Solve  the  following  equations,  taking  account  of  both  positive 
and  negative  square  roots  of  the  quantities  under  the  radical  sign. 
Apply  the  test  for  a  solution  in  each  case. 

1.  V(^-3)  =  3.  •     14.  y/{9  +  ^x)=2x-S. 

2.  V(»^-7)  =  4.  15.  Sx=^-{-y/(30x-n). 

3.  y/(iSx+l)=4.  16.  2  a:- 5  V^- 3. 

4.  5v'(x  +  7)=:4V(3a;-2).  17.  x -\- 3  +  ^(x  +  S)  =  6. 
6.  3y/(x  +  Z)=2y/(Sx  +  6).  18.  2x  +  1  =  v'C^  a;  +  3). 

6.  ^(x  +  5)=2y/(Sx  +  4).  19.  7a:  =  V(3a;-ll)+33. 

7.  3V(aJ  +  7)=5v(3x-2).  20.  y/(x+ 10)  +  y/(^x  +  1)=1. 

8.  y/(x  +  2)  =  x.  21.  ^x  +  y/{bx+l)=6. 

9.  y/{x-\-20)  =  x.  22.  ^x  +  y/{6x+l)  =  2. 

10.  x  +  y/(x  +  l)=6.  23.  V(2^  +  9)-V(«  +  4)=l. 

11.  x-y/(x  +  2)  =  4t.  24.  V(7a:+l)-V(3^+10)  =  l. 

12.  x-2  +  S^(x-2)  =  0.  25.  V(2a;+ll)-V(2a;-5)=2. 

13.  x-6  +  2^(x-b)=0.  26.  V(4a;  +  l)-V(a^  +  3)  =  2. 

27.  V(8a^  +  5)-2V(2a:  +  l)=l. 

28.  V(a^  +  4)  +  v'(^+20)-2V(a;  +  ll)  =  0. 

29.  V(4a;+l)-V(a^  +  3)  =  >/(»5-2). 


QUADRATIC   EQUATIONS.  249 

80.    V(2a;  +  4)  +  v(3x  +  7)  =  V(12a;  +  9). 


31.    V(6aJ  +  l)  +  v2(l-x)  =  V(7a;  +  6). 


32.  2V(2x+3)-V(5x  +  l)  =  V2(x-l). 

33.  V(x  -  l)(x  -  2)  +  V(x  -  3)(x  -  4)=y2. 

34.  v(a;+l)+_^_^  =  2. 
86.  y/(ix+l)  +  ^x  ^ 


V(l+^) 


36..      y/iS  +  X)-\-y/Xz=-^. 
y/X 

38.  V(«-^)  +  V(^ -a;)  =  \/(«  +  ^-2x). 

39.  y/iax  +  62)  +  ^(6x  +  a2)  =  a  _  5. 

40.  V(«  +  a;)  +  V(&  +  «)  =  \/(«  +  ft  +  2x). 

41.  y/(a-x)  +  y/ib-x)  =  y/{2a-{-2b'). 

183.  Eelations  between  the  roots  and  the  coefficients  of  a 
quadratic  equation. 

We  have  seen  [Art.  122]  that  cc^  -\-px-\-q  can  always 
be  expressed  in  the  form  {x  —  a)  {x  —  p). 

We  have  also  seen  [Art.  128]  that  if 

a^  +  px  -}-  q  =  {x  -  a)  (x  -  /3), 
then  a  and  /8  are  the  roots  of  the  equation 

a^ -{- px -{- q  =  0. 
But,  if 

x"^  -\-px  -{-q  =  {x  —  a){x  —  I3)  =  x^  —  {a-^  P)x-{-  aft, 
we  have  «  -f-  /3  =  —p  and  ajS  =  q. 


250  QUADRATIC   EQUATIONS. 

Hence  in  the  equation 

Qi?  +pa;  +  g  =  0, 
fhe  sum  of  the  roots  is  —p  and  the  product  of  the  roots  is  q. 
The  equation  aa:^+bx-\-c=0  becomes,  on  dividing  by  a, 

a        a 

Hence,  from  the  above,  if  a  and  p  be  the  roots  of 

aa^  +  bx  -\-  e  =  Oj 

b  c 

we  have  a-\-  3  = and  a3  =  -• 

a  a 

The  above  relations  between  the  roots  of  a  quadratic 
equation  and  the  coefficients  of  the  different  powers  of 
the  unknown  quantity  are  of  great  importance. 

Analogous  relations  hold  good  for  equations  of  the 
third  and  of  higher  degrees.  [See  Treatise  on  Algebra, 
Art.  129.] 

These  relations  may  also  be  deduced  as  follows : 

We  have  found  in  Art.  177  that  the  roots  of 

ax^  +  6a;  +  c  =  0 

b       ^(b'-4.ac) 

are  — o— ±         o -' 

2a  2a 

Call  these  roots  a  and  p  respectively ; 
then  «  =  __+Vi-^- ^ 

^         2a  2a 


QUADRATIC   EQUATIONS.  251 

By  addition  and  by  multiplication  we  then  have 

a 

Again,  the  quotient  of  a-{-  (3hj  ap  is 

a+^^1      1__6J 
a/3        a'^  (3  c' 

and  observing  that 

.  /I      1\2      2       1       1 

we  find  a^4-/3^  =  ^^_2_c  =  ^!^^, 

a^      a  or 

184.  Special  Forms.  If  one  or  more  of  its  coefficients 
vanish,  the  quadratic  equation  and  the  formula  for  its 
roots  become  simplified. 

(i.)  If  c  =  0,  the  equation  reduces  to 
X  {ax  -f  6)  =  0, 
the  roots  of  which  are  0  and  —  b/a. 

(ii.)  If  c  =  0  and  also  6  =  0,  the  equation  becomes 
ax^  =  0,  both  roots  of  which  are  zero. 

(iii.)  If  6  =  0,  the  equation  reduces  to 
aa^  +  c  =  0, 


252  QUADRATIC   EQUATIONS. 


the  roots  of  which  are  -\-^—c/a  and  —^—c/a.  The 
roots  of  a  quadratic  equation  are  therefore  equal  and 
opposite  when  the  coefficient  of  x  is  zero. 

(iv.)  If  a,  6,  and  c  are  all  zero,  the  equation  is  clearly 
satisfied  for  all  values  of  x. 

(v.)  The  case  in  which  either  a  =  0,  or  a  =  0  and 
6  =  0  (c  not  zero),  is  best  studied  through  the  general 
expressions  for  the  roots  of  asi?  -\-hx-\-  c  =  0,  but  in  the 
changed  form 

2c 2c 

That  these  expressions  are  respectively  equal  to 
_5  4-^(62_4^c)     -6-V(62-4ac) 
2a  '  2a 

follows  from  the  identity 

5  _  5  4-  ^(62  -  4ac)  IS  -  6  -  V(^'  -  4ac)  |  =  4ac. 

If  now  a  =  0,  one  root  becomes  —  c/6,  and  the  other 
2c/0.     And  if  a  =  0  and  6  =  0,  both  roots  become  2c/0. 

A  fraction,  such  as  2  c/0,  whose  numerator  is  a  finite 
quantity  and  whose  denominator  is  zero,  was  called,  in 
Art.  168,  an  infinite  quantity,  or  infinity,  and  was  denoted 
by  the  symbol  oo.     Thus  2c/0  =  oo. 

In  accordance  with  this  definition  and  with  the  results 
here  obtained,  when  it  becomes  necessary  to  interpret 
the  equations 

0.a;2  +  6a;  +  c=0, 

0.a^  +  0.a;  +  c  =  0, 


QUADRATIC   EQUATIONS.  263 

we  say  that  the  former  has  one  infinite  root,  the  latter 
two  infinite  roots.  Such  equations  present  themselves 
in  investigations  in  analytic  geometry. 

Ex.  1.    Solve  the  equation 


Vax  +  1  +  Vx  +  4  =  3. 
Squaring  both  members,  we  have 


(a  +  l)x  +  5  +  2\/(ax+  l)(x  +  4)  =  9; 

then  transposing  the  rational  terms  to  the  right  side  and  squaring 
again,  we  obtain 

4(ax  +  1) (X  +  4)  =  16  -  8(a  4-  l)x  +  (a  +  1)^x2, 

whence,  by  the  proper  reductions  and  transpositions, 

(a  -  1)2x2  -  12(2  a  +  l)x  =  0. 

Hence,  one  root  is  x  =  0,  and  for  this  one  both  square  roots 
must  be  taken  positively.    A  second  root  is 

12(2  g+l) 

and  it  requires  \/x  +  4  to  be  taken  negatively.  ^ 

If  a  =  1,  the  coefficient  of  x2  is  zero,  and  the  second  root  is 
infinite. 

Ex.  2.   Solve  the  equation 

y/(ax)-{-y/(x  +  l)  =  y/b, 
and  determine  for  what  value  of  b  one  root  is  zero,  and  for  what 
value  of  a  one  root  is  infinite. 

185.  Equations  with  given  Koots.  Although  we  cannot 
in  all  cases  find  the  roots  of  a  given  equation,  it  is  very 
eas3^  to  solve  the  converse  problem,  namely,  the  problem 
of  finding  an  equation  which  has  given  roots. 

For  example,  to  find  the  equation  whose  roots  are  4  and  5. 
We  are  to  find  an  equation  which  is  satisfied  when  x  =  4,  or 


254  QUADEATIC   EQUATIONS. 

when  X  =  5  ;  that  is,  when  x  —  4  =  0,  or  when  x  —  5  =  0  ;  and  in 
no  other  case.     The  equation  required  must  be 

(a;_4)(x-5)=0; 

for  this  is  an  equation  which  is  a  true  statement  when  x  —  4  =  0,  or 
when  aj  —  5  =  0,  and  in  no  other  case. 

If  we  get  rid  of  the  brackets  by  multiplying  out,  the  equation 
becomes 

x2  -  9  X  +  20  =  0  * 

Again,  to  find  the  equation  whose  roots  are  2,  3,  and  —  4. 

We  have  to  find  an  equation  which  is  satisfied  when  x  —  2  =  0, 
or  when  x  —  3  =  0,  or  when  x  +  4  =  0,  and  in  no  other  case.  The 
equation  must  therefore  be 

(a;_2)(x-3)(x  +  4)  =  0, 
that  is  x3  -  x2  -  14  X  +  24  =  0. 

Similarly,  the  equation  whose  roots  are  0,-1,  and  —  ^  is 
x(x+l)(a;+D=0. 
that  is  x8  +  fx2+ix  =  0. 

Ex.  1.  Find  the  equation  whose  roots  are  the  squares  of  the 
roots  of  the  equation  x^  +  5  x  —  7  =  0. 

Let  a,  /3  be  the  roots  of  the  given  equation ;  the«i  a^,  ^  will  be 
the  roots  of  the  required  equation.     Hence  the  required  equation  is 

(x-a2)(x-/32)  =  0, 

that  is  x2  - (a2  + /32)x  +  a2/32  =  0 (i.) 

*  The  equation  a;2  —  9  x  +  20  =  0  is  certainly  an  equation  with  the 
proposed  roots ;  but  to  prove  that  it  is  the  only  equation  with  the  pro- 
posed and  with  no  other  roots  we  must  assume  that  every  equation  has 
a  root. 

If  the  equation  x^ +  7x^  —  2=^0,  for  example,  had  no  roots,  then 
(x  —  4)  (x  —  5)  (x5  +  7  x2  —  2)  =  0  would  be  an  equation  vnth  the  pro- 
posed roots  and  with  no  others. 

The  proof  of  the  proposition  that  every  equation  has  a  root  is  too 
difficult  for  an  elementary  book. 


QUADRATIC  EQUATIONS.  255 

We  have  therefore  to  find  a^  +  ^  and  a^p^.  Now,  by  Art.  183, 
we  have 

a+/3=-5, 

and  a)8  =  —  7. 

Hence  a2  + /32  =(a  +  i3)2 -2a/3  =(- 5)2  -  2(- 7) 

=  25  4-  14  =  39 ; 
also  a2/32  =  49. 

Substituting  in  (i.),  we  have  the  required  equation,  namely 
a:2_39x  +  49  =  0. 

Note. — "We  might  have  obtained  the  required  equation  by 
finding  the  roots  of  the  given  equation,  and  squaring  them ;  but 
it  is  best  to  use  the  relations  found  in  Art.  183. 

Ex.  2.   If  a,  /3  are  the  roots  of  the  equation 

ax^ -h  bx -\-  c  =  0, 

find  the  equation  whose  roots  are  -  and  ^. 

/3  « 

The  required  equation  is 


(-|){.-!)-. 
(i*i} 


that  is  x2-p  +  tl  x  +  l=0. 

Now  -  +  §  =  ?1±^., 

/S      a  a/3      ' 

and,  by  Art.  183,  we  have 

a/3  =  ^anda2  +  /32  =  ^_2c 
a  a^      a 

...    tt^  +  /32^62       ^ 

*  *      a/3         ac       ' 
Hence  the  required  equation  is 

«2-(^_2^a;  +  l  =  0, 
or,  multiplying  by  ac, 

acx'^  -  (62  _  2  ac)x  +  ac  =  0. 


256  QUADRATIC   EQUATIONS. 

Ex.  3.   Show  that  no  mimerical  value  of  x  will  make  aj^— 4x+  5 
zero  ;  and  find  its  least  value. 

Since  a:2  -  4  x  +  5  =  (a;  -  2)2  +  1, 

and  {x  —  2)'^  is  always  positive,  \-\-{x  —  2)2  is  always  greater  than 
1,  except  when  a;  —  2  =  0,  and  then  it  is  equal  to  1. 

Thus  x2  —  4  a;  +  5  can  never  be  zero,  and  its  least  value  is  1. 

EXAMPLES   LV. 
Write  down  the  equations  whose  roots  are 

8.  0  and  -  4. 

9.  5,-3  and  0. 

10.  V2  and  -  V2. 

11.  ^5  and  -  y'S. 

12.  0,  ^3  and  -  v3. 

13.  2  -  V3  and  2  +  ^3. 

14.  5-^7  and  5  +  ^7. 
15.   a  —y/h  and  a  +  ^6. 

16.  Write  down  the  product  of  the  roots  of  each  of  the  follow- 
ing equations : 

(i.)  cc2  +  4x+ 1=0,  (iii.)  3a:2  +  5x-7  =  0, 

(ii.)  ic2  +  7 X  -  2  =  0,  (iv.)  5x2  -  x  -  1  =  0,    ■ 

(v.)  9ax2  +  36x  +  4c  =  0. 

17.  Write  down  the  sum  of  the  roots  of  each  of  the  following 
equations : 

(i.)  x2-4a2  =  o,  (iii.)  x2-5x  +  3  =  0, 

(ii.)  x2  +  3x  -  5  =  0,  (iv.)  2 x2  -  x  -  7  =  0, 

(v.)  6ax2  +  77>x  +  8  =  0. 

18.  Show  that,  if  a,  ^  be  the  roots  of  ax2  +  6x  +  c  =  0,  then 

1^1=  V(&2-4ac) 
o~/3  c 


1. 

2  and  -  2. 

2. 

3  and  -  4. 

3. 

-  3  and  -  2. 

4. 

\  and  —  |. 

5. 

\  and  -  1 . 

6. 

-  i  and  -  \. 

7. 

0  and  3. 

^ 


QUADRATIC  EQUATIONS.  257 

19.  Show  that,  if  a,  /3  be  the  roots  of  the  equation  ax^+bx+c=0, 
then 

(i.)  aB  +  ^3  =  ifcftf,        (i,)  1^  +  i      3«6c_^s, 

(iii.)  ^+^'  =  ^1=11^^  +  2. 

20.  Find  the  sum  of  the  squares  of  the  roots. of  the  equation 

x2  +  4  a;  +  2  =  0. 

21.  Find  the  sum  of  the  squares  of  the  roots  of  the  equation 

x2  +  ipx  +  p2  _  0, 

22.  Show  that  the  sum  of  the  squares  of  the  roots  of  the  equa- 
tion x2  +  ax  +  &  =  0  is  the  same  as  the  sum  of  the  squares  of  the 
roots  of  the  equation 

x2  +  3ax  +  b  +*4a2  =  0. 

23.  For  what  value  of  a  are  the  roots  of  the  equation 

equal  to  one  another  ? 

24.  For  what  values  of  a  are  the  roots  of  the  equation 

4x2'+(l  +  a)x+ 1  =0 
equal  to  one  another  ? 

26.  Find  the  value  of  a  in  order  that  one  of  the  roots  of 
100  x2  +  60  X  +  a  =  0  may  be  double  the  other. 

26.  Show  that  one  of  the  roots  of  x^  -{- px  -\-  q  =  0  ia  double  the 
other,  if  9^  =  2p2, 

27.  Show  that,  if  a,  /3  be  the  roots-  of  2  x^  -  5  x  +  3  =  0,  the 
equations  whose  roots  are  -  and  ^  is6x2—  13x  +  6  =  0. 

28.  Show  that,  if  a,  /3  be  the  roots  of  2x2  —  15x  +  4  =  0,  the 

equations  whose  roots  are  -  and  ^  is  8  x^  —  209  x  +  8  =  0. 
/3  a     ' 


258  QUADRATIC  EQUATIONS. 


29.  Form  the  equation  whose  roots  are  a  +  j8  and  -  +  -,  where 

a       ^ 
a,  jS  are  the  roots  of  x2  -  11  ic  +  22  =  0. 

30.  If  a,  j3  be  the  roots  of  a;^  ^  7^;  -|-  9  =  0,  find  the  equation 
whose  roots  are  "  "*"  ^  and  "  "^  ^. 

31.  Show  that  the  roots  of  x^ -(p"^ -2q)x  +  q- =  0  are  the. 
squares  of  the  roots  oi  x'^  -\-  px  +  q  =  0. 

32.  Find  the  condition  that  the  roots  of  a^x^  4-  6%  +  c^  =  0  may- 
be the  squares  of  the  roots  of  ax^  -\-  bx  -{-  c  =  0. 

83.   Show  that,  if  a,  /3  are  the  roots  of  px^  +  gx  +  r  =  0,  the  equa- 
tion whose  roots  are  a  +  j3  and  _f^£_  is  pqx^  +  {pr -{■  q^)x  +  qr  =  0. 

a  +  /3 

34.  Show  that,  if  a,  /3  are, the  roots  of  ax^  +  6ic  +  c  =  0,  the 

equation  whose  roots  are  a/3  and  —  is  acx^  —  (a^  -\-  c^)x  +  ac  =  0; 

a/3  1 

show  also  that  the  equation  whose  roots  are  a  +  /?  and  ^,   a  is 

a&a;2  +  (a^  +  &2)a;  _,.  ^5  _  0. 

35.  Show  that,  if  the  difference  between  the  roots  oix^+ax+b=0 
be  equal  to  the  difference  between  the  roots  of  x"^  -]- px -\-  q  =  0,  then 

a2_46=;)2_4g. 

36.  Show  that,  if  a,  /3  be  the  roots  of  x"^  +  px  +  q  =  0,  the  equa- 
tion whose  roots  are  (a  +  /3)2  and  (a  —  j8)2  is 

a;2  _  2(^2  _  2  g)a;  +^2(p2  _  4  g)  =  0. 

37.  If  a,  /3  be  the  roots  of  x^  -{-  px  +  q  =  0,  show  that  the 

equation  whose  roots  are  a  +  -  and  /S  +  -  is 
/3  a 

gx2 +p(l  +  g)a;  +  (1  +  g)2=  0. 

38.  Show  that,  if  a,  /S  be  the  roots  of  ax^  -j- bx -{■  c  =  0,  the 
roots  of  the  equation  (2  b^  +  ac)x^  +  3  abx  +  a^  =  0  are 

and 


a  +  2j8  jS  +  2a 


QUADRATIC   EQUATIONS.  259 

39.  Find  the  relation  that  must  exist  between  k  and  the  coeffi- 
cients of  the  equation  ox^  +  6x  +  c  =  0,  in  order  that  one  root  may 
be  k  times  the  other. 

40.  Determine  the  relation  that  must  exist  between  a  and  6,  in 
order  that  one  of  the  roots  of  the  equation 

y/{ax)  +  y/{x  +  a)  =  ^{x  +  h) 

may  be  zero,  the  value  of  a  that  will  make  one  of  its  roots  infinite, 
the  value  of  a  that  will  make  both  of  its  roots  infinite,  and  the 
relation  between  a  and  h  that  will  make  its  two  roots  equal. 


260  EQUATIONS   OF   HIGHER  DEGREE. 


CHAPTER   XVII. 

Equations  of  Higher  Degree  than  the  Second. 

186.  It  is  beyond  the  range  of  this  book  to  show  how 
to  solve  equations  of  higher  degree  than  the  second,  when 
the  equations  are  in  their  most  general  forms;  we  give 
however  some  easy  examples  of  such  equations. 

Ex.  1.     Solve  x*  -  6  cc2  4-  8  =  0. 

Here  the  equation  contains  only  two  powers  of  the  unknown 
quantity,  one  of  which  is  the  square  of  the  other ;  we  therefore 
proceed  as  in  Art.  176. 

We  have  ic*-6a;2 +  9-9+8  =  0, 

that  is  (x2  _  3)2  _  1  =  0. 

Hence         {(^2  -  3)  -  1}  {(x^  _  3)  +  1}  =  0  ; 
.  ••  a;2  —  4  =  0,  giving  x  =  ±  2, 
or  x2  —  2  =  0,  giving  x  =  ±  ^2. 

Thus  there  are  four  roots,  ±2,  ±^2. 

Ex.  2.     Solve  (x2  +  a;)2  +  4(x2  +  x)  -  12  =  0. 

In  this,  as  in  the  former  example,  the  unknown  quantity  occurs 
in  two  terms,  one  of  which  is  the  square  of  the  other.  In  all  such 
cases  we  can  bring  all  the  terms  over  to  the  left  side  of  the  sign  of 
equality,  and  then  resolve  the  expression  on  the  left  into  two  fac- 
tors as  in  the  preceding  chapter. 

In  the  present  instance  the  factors  can  be  seen  by  inspection. 

For  ^2  _|.  4^  _  12  =(^  +  6)(^  -  2); 

therefore  the  equation  may  be  written 

{(x2  +  x)+6}{(x2  +  x)-2}  =  0. 


EQUATIONS   OF   HIGHER  DEGREE.  261 

Hence  a;2  +  x  +  6  =  0,  or  x^  +  a;  -  2  =  0. 


The  roots  of     a;2  ^  x  +  6  =  0  are  -  i  ±      ^^- 

The  roots  of     x^  +  a;  -  2  =  0  are  1  and  —  2. 
Hence  the  given  equation  has  the  four  roots 


1,     -2,     -\±\yr^^. 

Ex.3.     Solve  -^  +  ^±2  =  37. 

X  +  2        x2         6 
Here  one  of  the  terms  in  which  x  occurs  is  the  reciprocal  of  the 

/J.2 

other.    If  we  put  y  = ,  we  find  y  from  the  quadratic  equation 

y      6 
or,  multiplying  by  6  y, 

6  ?/2  _  37  y  -(-  6  =  0, 

that  is  (6  y  -  1)  (y  -  6)  =  0. 

Hence  y  =  |,  or  y  =  6. 

Thus  ^_=1,  or-^=6. 

a;  +  2     6        x  +  2 

In  the  first  case,  we  have  6  x^  —  x  —  2  =  0,  the  roots  of  which 
are  \  and  —  \. 

In  the  second  case,  we  have 

a;2  _  6  X  -  12  =  0, 

the  roots  of  which  are  3  ±  V^l- 

Hence  the  given  equation  has  the  four  roots, 

I,     -I,    3±V21. 

Ex.  4.     Solve  4x2-6x  +  3v(2x2-3x+  7)  =  30. 

Equations  of  this  form,  in  which  the  ratio  of  the  coefficients 
of  x2  and  x  in  the  terms  under  the  radical  is  equal  to  the  ratio  of 
the  coefficients  of  x2  and  x  in  the  terms  outside  the  radical,  are  not 
uncommon. 


262  EQUATIONS   OF   HIGHER  DEGREE. 

.  To  solve  the  equation,  put 

V(2x2-3x+7)=y, 
then  2  x2  -  3  a;  +  7  =  y2, 

2  a;2  -  3  a;  =  2/2  _  7^ 
and  4x2-6ic  =  2  2/2-14. 

Hence,  from  the  given  equation,  we  have 
22/2 -14 +  32/ =  30; 
that  is  2  2/2  +  3  2/  -  44  =  0, 

that  is  (2/-4)(2  2/+ 11)=0; 

hence  2/  =  4,  or  y  =  —  y^ ; 

.  •.  2/2  =  16,  or  2/2  =  if  i. 
Since  2/2  =  2  a;2  _  3  a;  _|_  7^  we  have 

I.  2x2-3x  +  7  =  16, 
that  is  2  x2  -  3  x  -  9  =  0, 
that  is                              (2  X  +  3)  (x  -  3)  =  0  ; 

.  %  X  =  3,  or  X  =  —  f . 

II.  2x2-3x+7  =  4J-, 
that  is                               2  x2  -  3  X  -  -9^  =  0, 

the  roots  of  which  are  -  ±  :s/}^. 
4         4 

187.  It  has  been  shown  [Art.  148]  that  if  a  rational 
and  integral  expression  containing  x  vanishes  when  any 
particular  value  a  is  given  to  a?,  then  a;  —  a  is  a  factor  of 
that  expression. 

For  example,  x^  —  7  x  +  6  vanishes  when  2  is  put  for  x  ;  there- 
fore, by  the  above  theorem,  x  —  2  is  a  factor,  and  by  division  we 
find  that 

x3  -  7  X  +  6  =  (X  -  2)  (x2  +  2  X  -  3) . 

Again,  x^  —  4x2  +  2x+l  vanishes  when  x  =  1 ;  therefore  x  —  1 
is  a  factor,  and  by  division  we  find  that 

ic3  _  4x2  +  2x  +  l  =  (x  -  l)(x2  -  3x  -  1). 


EQUATIONS   OF   HIGHER   DEGREE.  263 

188.  It  follows  from  the  theorem  enunciated  in  the 
last  article  that  when  one  root  of  an  equation  is  known, 
the  degree  of  the  equation  can  be  lowered. 

By  the  application  of  this  principle,  the  roots  of  the 
important  equation  oc^  —  1  =  0  can  be  found. 

Ex.  1.   Solve  the  equation  x^  —  1  =  0. 

Since  a^  -  1  ={x  -  l^^x"^  +  x  +  1), 

we  have  (x  -  1)  (x^  +  x  +  1)  =  0. 

Hence  x  =  l; 

or  else  x^  +  x  +  I  =  0, 

the  roots  of  which  are  —  a  ±  \/(~  !)• 

Hence  there  are  three  roots  of  the  equation  a^  =  1,  one 
being  real  and  the  other  two  imaginary.  Thus  there  are 
three  quantities  whose  cubes  are  equal  to  1 ;  that  is,  there 
are  three  cube  roots  of  1,  which  are 

1.  -i  +  V(-f)  and  _^-V(-f). 

[See  Treatise  on  Algebra,  Art.  139.] 

Ex.  2.  One  root  of  the  equation  x^— 7x  +  6  =  0  is  2;  find  the 
other  roots. 

Since  x^  —  7  x  +  6  vanishes  when  x  =  2,  x  —  2  must  he  a  factor, 
and  by  division  we  find  that 

x8  -  7  X  +  6  =  (X  -  2)  (x2  +  2  X  -  3) . 

Hence  (x  -  2)(x2  +  2x  -  3)  =  0. 

Hence  the  other  roots  of  the  equation  are  those  given  by 

x2  +  2x-3  =  0, 

that  is  (x  +  3)(x-l)  =  0. 

Thus  the  cubic  equation  x^  —  7x  +  6  =  0  has  the  three  roots  1, 
2,  and  -  3. 


264  EQUATIONS   OF   HIGHER   DEGREE. 

Ex.  3.   For  what  values  of  x  will 

0-4-25x2  + 30a; -9, 

and  X*  -  8  ic3  +  19  a;2  -  14  x  +  3 

both  vanish  ? 

If  both  expressions  vanish  for  the  s^me  value  of  x,  say  cc  =  a, 
they  will  both  ha^^e  a;  —  a  as  a  factor.  Now  we  can  find  the  com- 
mon factor  of  any  two  expressions  by  the  ordinary  process  of 
finding  their  H.  C.  F. 

In  the  present  case,  the  H.  C.  F.  will  be  found  to  be 

x2  -  6  X  +  3  ; 

and  since  ic2  —  6  x  +  3  is  a  factor  of  both  expressions,  and  is  their 
highest  common  factor,  both  expressions  will  vanish  for  the 
values  of  x  given  by 

x2-5cc  +  3  =  0, 

and  for  no  other  values.     Thus  the  values  required  are  the  roots  of 

a;2  -  5  X  +  3  =  0, 


and  these  roots  are 


5  ±  V13, 


EXAMPLES  LVI. 
Find  the  roots  of  the  following  equations  : 


c4  _  2  x2  -  8  =  0. 


a^ 


4.   X*- 7x2 -18=0.  10.  (x2-x)2-8(x2-x)  +  12=0. 

5    x2-l  ^   1  _i^Q  11-  (x2+x)2-22(x2+x)  +  40zz0. 

9         ^'^  12.  (x2  +  x)(x2  +  x+l)=42. 

6.  x2  +  l^  =  29.  13.  _^  +  ^+l  =  2. 

^^  X  +  1  X2 

7.  x2  +  l=a2  +  l.  14.  ^_+?i±I  =  ^. 
•^  ^a;2             a2  x2+l^     x         2 


EQUATIONS   OF   HIGHER   DEGREE.  265 

16.    _^+£^  =  lI. 
x-1        x^         4 

16.    (x2  +  x+l)(a;2  +  x-f)+l  =  0. 

17.  (xHa;+l)(x2+x+2)  =  12.        2Q    ^.  ,  ^  ,  ^  ^     42 

18.  (.+  1)V4(.  +  1)  =  12, 


a;2  +  x 


21.   2x2-4 a;-Va;2-2x-3=9. 

19         x2+2        x2+4x+l_5 

•    X24-4X+1        X24-2      ~2         22.   x2+ Vx2+3x+7=23-3x. 

23.  2x2  4-6x  =  l-V(aj2  +  3a;  +  i). 

24.  x2  +  v(4x2  +  24x)=24-6x. 

25.  2(2x-3)(x-4)-V(2a;2-lla;+15)=60. 

26.  x2  +  (X  -  2)  (X  -  3)  +  V2  x2  -  5  X  +  6  =  6. 

27.  (x+6)(x-2)-36=V(x  +  4)(x-l). 

Solve  the  following  cubic  equations  having  given  one  of  the  roots 
of  each : 

28.  x8-2x+  1  =0,  [x=  1]. 

29.  x8-6x  +  4=:0,  [x  =  l]. 

80.  x8-49x  +  120  =  0,  [x  =  3]. 

81.  x8  -  3x2  -  7 X  +  21  =  0,  [X  =  3]. 

82.  x8-2x2-7x-4  =  0,  [x  =  -l]. 
88.  5x8  -  16  x2  +  3  X  +  14  =  0,  [x  =  2]. 

34.  For  what  values  of  x  will  x*  +  2  x2  +  9  and  x*  -  4  x  +  16 
both  v&,nish? 

35.  The  equations 

2x8  +  21x2 +  34x- 106  =  0, 
2x3-      x2-76x- 105  =  0 
have  one  root  in  common  ;  find  it. 
Solve  the  following  equations : 

36.  x8-21x  +  20  =  0.  39.   x8  -  3x2  -  60x  -  100  =  0. 

37.  x8+2x2-9x-18=0.  40.   x3-19x  +  30  =  0. 

38.  x8-x2-22x+40=0.  41.   2x8  +  7x2  -  5x  -  4  =  0. 

42.   4x8 +4x2- 9x- 9  =  0. 


266  SIMULTANEOUS  EQUATIONS. 


CHAPTER  XVIII. 

Simultaneous    Equations    of   the   Second 
Degree. 

189.  We  now  proceed  to  consider  simultaneous  equa- 
tions, one  at  least  of  wMcli  is  of  the  second  degree. 

We  first  take  the  case  of  two  equations  which  contain 
two  unknown  quantities,  one  equation  being  of  the  first 
degree  and  the  other  of  the  second  degree. 

Example.     Solve  the  equations 

From  the  first  equation  we  have 

X  =  6  —  2y. 

Substituting  this  value  of  x  in  the  second  equation,  we  have 

(5_22/)2  +  22/2  =  9; 

.-.  6y^-20y  +  16  =  0; 

.-.  32/2- 102/ +  8  =  0; 

that  is  (3  y  -  4)  (?/  -  2)  =  0. 

Hence  ?/  =  |,  or  y  =  2. 

If  2/  =  t,  x  =  6-2y  =  S-^  =  i; 

and  if  2/  =  2,   a;  =  5  —  2?/  =  5  —  4  =  1. 

Thus  we  have     x  =  |,   y  =  § ',   or  x  =  1,   y  =  2. 

[The  result  should  not  be  written  in  the  ambiguous  form  a:  =  |, 
orl;  2/  =  I,  or2.J 


SIMULTANEOUS   EQUATIONS.  267 

From  the  above  example  it  will  be  seen  that  two 
equations,  one  of  which  is  of  the  first  degree  and  the 
other  of  the  second  degree,  can  be  solved  in  the  following 
manner: 

From  the  equation  of  the  first  degree  find  the  value  of 
one  of  the  unknown  quantities  in  terms  of  the  other  unknown 
quantity  and  the  known  quantities,  and  substitute  this  value 
in  the  equation  of  the  second  degree :  one  of  the  unknown 
quantities  is  thus  eliminated,  and  a  quadratic  equation  is 
obtained,  the  roots  of  which  are  the  values  of  the  unknown 
quantity  which  is  retained, 


Ex.  1.    Solve  the  equations 

Sx  +  Ay  =  5,  2x^  -xy  +  y^  =  22. 

From  the 

first  equation,  x  =     ~     ^-     Hence,  substituting  in 

the  second,  we  have 

K^T-r-^^)-^--' 

which  reduces  to 

63  y2_95y_  148  =  0, 

that  is 

(y+l)(53y-148)  =  0. 

Hence 

— -  =  f- 

If 

v  =  -i,.  =  411  =  3. 

And,  if 

148           6-4y          109 
^53'               3                53 

Ex.  2.  Solve  xy-hx  =  25,  2xy-3y  =  2S. 
Multiply  the  first  equation  by  2,  and  subtract  the  second ;  then 
2  X  +  3  2/ =  50  -  28  =  22. 


268  SIMULTANEOUS  EQUATIONS. 

H«nce  y  =  ^^: 

and  substituting  in  the  first  equation,  we  have 


f22-2x\ 


+  a;  =  25  ; 


.-.  2x2 -25a: +  75  =  0, 
whence  we  have  x  =  5  or  x  =  Y-,  the  corresponding  values  of  y 
being  4  and  |,  respectively. 

Ex.3.  Solve 

2«2_3aj-42/  =  47, 

Multiply  the  second  equation  by  2  ;  then 

6x^-\-Sx  +  iy  =  ns\ 
Adding  the  last  equation  to  the  first,  we  ha^e- 
8x2  + 5a;  =  225, 
whence  x  =  5  or  x  =  —  -*/,    The  values  of  y  are  then  found  by  sub- 
stituting in  either  of  the  given  equations. 

EXAMPLES  LVII. 
Solve  the  following  equations : 

1.  X  +  y  =  6,  6.   X  +  2/  =  15, 

x2  —  j/2  =  24.  4:X  —  4:y  =  xy. 

2.  X  —  y  =  10,  7.   2  X  —  2/  =  5, 
x2  +  y2  =  58.  x  +  Sy  =  2xy. 

3.  3x+3y  =  10,  8.  3x-31  =  5y, 
xy=l.  x2  + 5x2/ +  25  =  2/2. 

4.  2x  +  3y  =  0,  9.   3x  +  2y  =  5, 

4x2 +  9x2/ +  92/2  =  72.  x^  -  4xy  +  5y^  =  2. 

5.  2x-52/  =  0,  10.   2x-72/  =  25, 

x2- 32/2=  13.    *  5x2 +  4x2/ +  32/2  =  23. 


2  +  3  =  6. 
X     y 

12. 

a;  +  2y  =  7, 

?  +  «  =  5. 

X     y 

13. 

a:  +  y  =  5, 

1  +  1  =  1 
X     y     Q 

14. 

x-y  =  l. 

x_y  _5 
y     X     6* 

15. 

2x-y  =  4, 

^4-^  =  1. 

X     y 

SIMULTANEOUS  EQUATIONS.  269 


31.   x  +  y  =  2,  16.   7x-3y  +  7  =  0, 

^-1^  =  6. 
X      y 

17.  xy  +  x=15j 
xy  —  x  =  8. 

18.  xy  +  2x=5f 
2xy-y  =  Z. 

19.  a;2-y  =  29, 
a;2  +  a:  +  y  =  49. 

20.  a;2  +  3a;-2y  =  4, 
2a;2-6x  +  3y  +  2  =  0. 


190.  It  should  be  remarked  that  the  methods  we  have 
thus  far  explained  do  not  enable  us  to  solve  any  two 
equations  which  are  both  of  the  second  degree ;  for  the 
elimination  of  one  of  the  unknown  quantities  will  fre- 
quently lead  to  an  equation  of  higher  degree  than  the 
second,  from  which  the  remaining  unknown  quantity- 
would  have  to  be  found ;  and  we  have  not  j^et  learned 
how  to  solve  an  equation  of  higher  degree  than  the 
second,  except  in  very  special  cases. 

For  example,  if  we  have  the  equations  x^  +  a;  +  y  =  3  and 
x2  +  2/2  =  5.  We  have  from  the  first  equation  y  =  3  —  x  —  a;^  ;  and, 
by  substituting  this  value  of  y  in  the  second  equation,  we  get 

X2  +  (3  -  X  -  X2)2  =  5, 

that  is  a4  +  2x8-4x2-6x  +  4  =  0; 

and  this  equation  of  the  fourth  degree  cannot  be  solved  by  any 
methods  which  are  within  the  range  of  this  book. 


270  SIMULTANEOUS   EQUATIONS. 

191.  We  can  always  solve  two  equations  of  the  second 
degree  when  all  the  terms  which  contain  the  unknown 
quantities  are  of  the  second  degree.  The  method  will  be 
seen  from  the  following  example. 

Ex.  1.     Solve  the  equations 

x?/  +  4  2/2  =  8. 

Divide  the  members  of  the  first  equation  by  the  corresponding 
members  of  the  second  equation  ;  we  then  have 

a;2  +  3  a;y  _  28  _  7^ 
o-y  _l_  4  y2  ~  8  ~  2' 

Hence  2(x2  +  3  x^)  =  7(a:!/  +  4  y2) ; 

.-.  2^2 -xy- 28  2/2  =  0, 

that  is  (2x  +  7?/)(ic-4  2/)  =  0. 

Hence  x  =  4  y,  or  x  =  —  |  ?/. 

I.  If  X  =  4  2/,  we  have  from  the  second  equation 

42/2,+  42/2  =  8;  .-.  y=±l. 
And  therefore         x  =  4 «/  =  ±  4. 

II.  If  X  =  — 12/,  we  have  from  the  second  equation, 

-12/2  +  4^2  =  8; 
.-.  y  =  ±4. 
And  therefore         x  =  —  1 2/  =  =F  14. 

Thus  there  are  four  sets  of  values,  namely,  x=4,  2/=l;  x  =  —  4, 
2/  =  —  1 ;  X  =  14,  2/  =  — 4  ;  and  x  =  —  14,  2/  =  4. 
We  have  in  the  above  written  down  the  factors  of 

2  x2  -  X2/  -  28  2/2 

by  inspection  :  when  this  cannot  be  done,  the  factors  can  be  found 
by  the  method  of  Art.  122. 


SIMULTANEOUS   EQUATIONS.  271 

Ex.  2.     Solve  x2  -  3x2/  =  0,  5x2  +  3^2  =  48. 
The  first  equation  is  x(x  —  3  y)  =  0. 
Hence  x  =  0,  or  x  =  3  y. 

If  X  =  0,  the  second  equation  gives  Sy"^  =  iS;   .*.  2/  =  ±  4. 
If  X  =  By,  the  second  equation  gives  45 ^24.3^2—48  j  . •.  y—  _j- 1^ 
and  then  x  =  ±  3. 

192.  Any  pair  of  equations  which  are  homogeneous,  so 
far  as  the  terms  which  contain  the  unknown  quantities 
are  concerned,  can  be  solved  by  the  method  adopted  in 
the  last  article.  Sometimes,  however,  the  equations  can 
be  solved  by  special  methods. 

For  example,  to  solve 

x2  +  y2  _  74^ 

2  «y  =  70. 
By  addition,  we  have 

x2  +  2  xy  +  2/2  =  144, 
that  is  (x  +  y)2  -  122  =  0  ; 

.-.  (x  +  2/-12)(x  +  2/  +  12)  =  0. 

Hence  x  +  y  =  l2 (1.), 

x  +  y=-12 .     .     (ii.). 

Again,  from  the  given  equations,  we  have  by  subtraction, 
aj2  -  2  xy  +  1/2  =  4^ 

that  is  (X  -  t/)2  -  22  =  0  ; 

.'.  (x-2/-2)(x-2/4-2)  =  0. 

Hence  x  —  y  =  2 Oii-)> 

x-y  =  -2 (iv.). 

From  (i.)  and  (iii.),  we  have  x  =  7,  y  =  S. 
From  (i.)  and  (iv.),  we  have  x  =  5,  y  =  7. 
From  (ii.)  and  (iii.),  we  have  x  =  —5,  y  —  —7. 
And,  from  (ii.)  and  (iv.),  we  have  x  =  —  7,  y  =  —  5. 


272  SIMULTANEOUS   EQUATIONS. 

Thus  there  are  four  pairs  of  values,  two  of  which  are  given  by 
aJ  =  ±  7,  ?/  =  ±  5,  and  the  other  two  hy  x  =  ±  5,  y  =±7,  it  being 
understood  that  in  both  cases  the  upper  signs  are  to  be  taken 
together,  and  the  lower  signs  are  to  be  taken  together. 

Again,  taking  the  equations  considered  in  Art.  191,  namely, 

x^  +  Sxy  =  28, 

«!/  +  4  ?/2  =  8. 

We  have  by  addition  x^  -\- 4:xy  -\-  iy^  =  36, 

that  is  (x  +  2  y)2  -  62  =  0  ; 

.-.  x  +  2y  =  6 (1.), 

or  x  +  2y  =  -6 (ii.). 

We  can  now  complete  the  solution  as  in  Art.  189,  by  taking 
(i.)  and  (ii.)  with  either  of  the  given  equations. 

EXAMPLES  LVIII. 

1.  a;2_2xy  =  0,  9.  x^-Sy'^  =  lS, 
4aj2  + 9i/2  =  225..  Sx^-y^  =  n. 

2.  2x2-3xy  =  0,  10.   Sxy  +  x^  =  10, 
y2  -I-  5xy  =  34.  5xy  -  2 x2  =  2. 

8.   x2  +  3x2/  =  45,  11.   x2  +  3x2/  =  54, 

y2_x2/  =  4.  xy  + 4^2  =  115. 

4.  x2_x?/  =  63,  12.  x(x  +  y)  =  40, 

y^   +    Xy    =   22.  y(^^_y^^Q^ 

5.  x2  +  a;?/  =  24,  13.   x^  -  3xy +  2y^  =  S, 
2  2/2  +  3xy  =  32.  2x^  +  y^  =  6. 

6.  x2-3xy  =  10,  14.  a;2-2xy  +  5  =  0, 
42/2-xi/  =  -l.  (a;_y)2_4  =  0. 

7.  x2  +  x?/-2y2  =  _44,  15.  x^  +  5y'^  =  S^, 

xy  +  3 2/2  =  80.  3x2  +  17 xy  -  y2  =  -  84. 

8.  x2  +  3x2/  =  7,  16.   x2-7x2/-9y2  =  9, 
y^  +  xy  =  6.  x2+ 6x2/  + 11 2/2  =  5. 


SIMULTANEOUS   EQUATIONS.  273 


17. 

4a;2-3a;y  =  10, 
y^-xy  =  6. 

23. 

3     2     3 

18. 

x^  +3xy  =  40, 

2-  +  ?  =  0. 

4y^+xy  =  9. 

19. 

x^  +  xy  +  y-i  =  7, 
6x^-2xy  +  y'^  =  6. 

24. 

2     5_6 
«     2/     6' 

20. 

2  x2  -  2  a;?/  +  3  2/2  =  18, 
3x2-2^/2  =  19. 

21. 

jc+2/  +  l=0, 

22. 

a:     2/      12 

a;  +  y  +  3  =  0, 

1  +  1-1  =  0. 
X     y     6 

25. 
26. 

(a;  +  l)(j/  +  l)=10, 
a;2/  =  3. 

(x  +  3)(2/  +  l)=4, 
xy  +  1  =  0. 

193.   The  following  examples  will  show  how  to  deal 
with  some  other  cases  of  simultaneous  equations. 

Ex.1.   Solve  x-y  =  2 (i.), 

x3- 2/8  =  386 (ii.). 

From  (i.),  we  have  x  =  y  +  2. 

Substitute  this  value  in  (ii.),  and  we  have 
(y  +  2)8  -  2/8  =  386, 
that  is  6  2/2  +  12  2/  +  8  -  386  =  0 ; 

.  ♦.  2/2  4-  2  2/  -  63  =  0. 
Hence  y  =  7,ory  =  —  9. 

If2/  =  7,  x  =  2/  +  2  =  9. 

If  2/  =  - 9,  x  =  j/+2  =  -7 

Thus  X  =  9,  2/  =  7,  or  X  =  —  7,  y  =  —  9, 

Ex.  2.  Solve  x-3y  =  2, 

x2  -  9  2/2  =  8. 


274  SIMULTANEOUS   EQUATIONS. 

Divide  the  members  of  the  second  equation  by  the  correspond- 
ing members  of  the  first ;  then 

X  +  3y  =4; 
and  from  x  +  3  y  =  4  and  aj  —  3  ?/  =  2, 
we  have  x  =  3,  y  —  \. 


Ex.  3.     Solve 

'  We  may  consider 
107,  Ex.  3. 

^and 

X 

1 

y 

as  the  unknown  qna,n titles,  as  in  Art. 

Square  both  members  of 

0 

:.)  ;  then 

l  +  ~  +  l  =  49 (111.). 

a;2     xy     y'^  ^     ^ 


From  (ii.)  and  (iii.),  we  have 


—  =  49-25  =  24 (iv.). 

xy 


From  (11.)  and  (iv.),  we  have 


1^A  +  1  =  25-24  =  1, 
x^     xy     y^ 


that  Is 


\x    y) 


.•.1-1=±1.    .    .   \    .    .     (v.). 
X     y 

The  values  of  -  and  -  are  at  once  found  from  (1.)  and  (v.). 
X         y 

Ex.4.    Solve  x2  -  x?/ +  ?/2  =  61 (1.), 

X4  +  ic2y2  +  y4  =  1281 (ii.). 

Divide  the  members  of  (11.)  by  the  corresponding  members  of 
(1.)  ;  we  then  have 

x'^  +  xy  -\-y'^  =  2l (Hi.). 


SIMULTANEOUS   EQUATIONS.  275 

From  (i. )  and  (iii. )  we  have,  by  subtraction, 

2xy  =  -  40, 
or  xy  =  -20      .....     .     (iv.). 

From  (i.)  and  (iv.),  we  have 

a;'-^- 2x^  +  2/2  =  81; 

.-.  a:-y  =  +9 (v.), 

or  x-2/  =  -9 (vi.). 

From  (1.)  and  (iv.),  we  have  also 

a;2  +  2a;y  + 2/2  =  1; 

.-.  x4-2/  =  l (vii.), 

or  x-\-y  =  —\ (viii.) . 

Then  combining  either  of  the  equations  (v.)  and  (vi.)  with 
either  of  the  equations  (vii.)  and  (viii.)  we  get  the  four  pairs  of 
values  x  =  ±5,  y=T4;  x  =  ±4,  y  =  q=6. 

Ex.6.    Solve  x2_2xy  =  3y (i.), 

2x2-9y2  =  9y (ii.). 

Multiply  both  sia'cc  of  (i.)  by  3  ;  then 

3x2-6xy  =  9j,-    .  ^.  (iii.). 

Hence  2x2  -  9y2  =  3x2 -65// ,         ^ 

.-.  x2-6xy +  9y2  =  o,  ^^^ 

that  is  (x  -  3  y)2  =  0 ; 

.-.  x  =  3y. 
Substitute  in  (11.)  ;  then 

2(3z/)2-92/2  =  9y, 
or  9  y2  _  9  y  =  0  ; 

.♦.  y(y-i)=o. 

Hence  y  =  0,  or  y  =  1. 

If2/  =  0,  x  =  3y  =  0. 

Ify  =  l,  x  =  3«/  =  3. 

Thus  X  =  0,  J/  =  0  ; 

or  else  x  =  3,  y  =  1. 


276  SIMULTANEOUS   EQUATIONS. 

Ex.  6.     Solve  X  -  ?/  =  2,  x^-y^  =  242. 

Put  x  =  z-\-l,  then  y  =  z  —  1. 

Hence 

x^-y^=(z  +  iy-(iz-iy 

=  z^+6  z^  +  10  z^+lO  z'^-\-6  z+1-  (z^-b  z^+lO  z^-10  z^+5  z-1) 

=:lOz*  +  20z^  +  2. 

Hence  10  z^  +  20  z"^ -\- 2  =  242  ; 

.'.  z*  +  2z'2  =  2^', 

Hence  z  =  ±2,  or  z  =  ±  V-  6. 

If  «  =  ±  2,  X  =  3  or  -  1,  and  ?/  =  1  or  —  3. 

It  z=±  ^/^^,  X  =  1  ±  V^^,  and  y=-l  i.V^^. 

Ex.  7.    Solve  x?/  +  x0  =  27 (i.), 

yz  +  yx  =  S2 (ii.), 

ZX  +  zy  =  S5 (iii')' 

From  (i.)  and  (ii.),  we  have  by  addition 

2xy  +  xz  +  yz-bQ     .■>'•;.    .     .     .    (iv.). 
From  (iv.)  and  (iii.),  we  have  by  subtraction 
2  xy  =  24  ; 

.:  xy  =  12 (v.). 

Hence,  from  (L),  x^  =  15 (vi.). 

And,  from  (ii.),  yz  =  20 (vii.)- 

From  (v.)  and  (vi.),  we  have  by  multiplication 

x'^fz  =  180 (viii.). 

From  (vii.)  and  (viii.),  we  have  by  division 
x2  =  180/20  =  9 ; 
.-.  x  =  ±3. 
Hence,,  from  (v.) ,    «/  =  12/  (  ±  3)  =  ±  4. 
And,  from  (vi.),      ^  =  15/(±  3)  =  ±  5. 
Thus  x  =  ±3,  y  =  ±4,  z  =  ±5. 

All  the  upper  signs  being  taken  together,  and  all  the  lower  signs 
being  taken  together. 


SIMULTANEOUS   EQUATIONS.  277 

EXAMPLES  LIX. 

1.  x-y  =  S, 
x3  _  ?/3  =  279. 

2.  x-y  =  2, 
x3  _  ?/3  =  98. 

3.  x  +  y  =  l, 
x3  +  1/3  =  91. 

4.  x-\-y  =  1, 

a^  +  yi  =  61. 

6.   x^  +  xy  +  y^  =  lS, 

Xi  +  xY  +  y4  =  91. 

6.  x^  —  xy  +  y^  =  9, 
x*  +  x2?/2  +  y4  =  243. 

7.  xy(x-\-y)  =  2^0, 
x^  +  y^  =  280. 

8.  xy(x-  y)=  12, 
x3  -  w3  =  63. 

19.   2x2-x2/  +  2/2  =  2y, 

«ll2  2x2  +  4xi/  =  5y. 


X     y 
1    .   1 


13. 

x-^y  =  l. 

x2y2  +  13  x?/  +  12  =  0. 

14. 

X  +  2/  =  5, 

4x?/  =  12  -xV-  ^ 

15. 

x2  -  xy  +  2/  =  -  6, 

y^-xy-x  =  12. 

16. 

x2  +  x?/  -  y  =  9, 

2/2  +  X2/  -  X  =  -  3. 

17. 

x4-i  =  3, 

18. 

^_l1      18 

y     7 

'-i=^- 

20.  x3  4-  1  =  9  y, 
;„     x2  +  y'2  =  20.  x2  +  x  =  6|/. 

21.  2x3 +  5 1/8  =  115, 

10.   ?  +  l  =  l,  3x3  +  7y8^186. 
X      1/ 

2  3       1  _  f.                         22.  x22/  +  a;?/2  =  180, 
^■^^"^^2-^-  x2y2  =  400. 

3  i_  23.  a^  +  x2y2  +  yt  =  91, 

"•     ^-^2-^'  (x2-Xl/  +  l/2)(x-l/)2  =  28. 

A_JL4.2__3^  24.  a;24-3x2/+2/2+4(x+y)=13, 

x2     xy     2/2       *  •3x2-xi/  +  3?/2+2(x+2/)=9. 

12.   1-A  =  3,  26.  X2/  +  ?  =  ^ 


4 1/2  y     3 

x2     xy     4i/2  ^^x     6 


278 


SIMULTANEOUS   EQUATIONS. 


X     xy      S 

y    3 


28.  a(x-a)=  b(y-b), 
xy  =  ax  +  by. 


29.  X = y  =  a 

y      X      ^ 


yz  =  4,  zx  =  9,  xy  =  16. 


27.  ax  =  by, 

(x  -  a)(y  -b)=  ab.  30. 

31.  x'^yz  =  6,  y'^zx  =  12,  ;s2a.y  ^  ig. 

32.  a;(x+?/  +  2!)  =  8,  ?/(x  +  y +2)=  12,  2;(sc  +  ?/ +  0)=  6. 

33.  x(y  +  z)=e,     y(z-j-x)=12,     z(ix  +  y)=  10. 

34.  (_x  +  y)(ix  +  z)=2,      (y  +  z)(y  +  x)=S,     {z -\- x) (z  +  y)  =  6. 

35.  y^;  =  a^,   zx  =  b^,   xy  =  c^. 


36. 


y±z_z-hx  _x 


=  2  xyx. 


a  b  c 

37.  (l/  +  5?)(a;  +  y +  ;2)=6  +  c, 

(^z  -\-  x)(x  +  y  +  z)=  c  +  a, 

(^  +  y)  (aj  +  w  4-  is;)  =  a  +  &. 


PROBLEMS.  279 


CHAPTER  XIX. 

Problems. 

194.  We  now  give  examples  of  problems  in  whicli  the 
relations  between  the  known  and  unknown  quantities  are 
expressed  algebraically  by  means  of  quadratic  equations. 

In  the  solution  of  problems  it  often  happens  that  by 
solving  the  equations  which  are  the  algebraical  state- 
mrnts  of  the  relations  between  the  magnitudes  of  the 
known  and  unknown  quantities,  we  obtain  results  which 
do  not  all  satisfy  the  conditions  of  the  problem. 

The  reason  of  this  is  that  the  roots  of  the  equation 
are  the  numbers,  whether  positive  or  negative,  integral  or 
frattional,  which  satisfy  it;  but  in  the  problem  itself 
there  may  be  restrictions,  expressed  or  implied,  on  the 
numbers,  and  these  restrictions  cannot  be  retained  in  the 
equation.  For  example,  in  a  problem  which  refers  to 
a  number  of  men,  it  is  clear  that  this  number  must  be 
integral,  but  this  condition  cannot  be  expressed  in  the 
equations. 

Thus  there  are  three  steps  in  the  solution  of  a  prob- 
lem. We  first  find  the  equations  which  are  the  algebrai- 
cal expressions  of  the  relations  between  the  magnitudes 
of  the  known  and  unknown  quantities ;  we  then  find  the 
values  of  the  unknown  quantities  which  satisfy  these 
equations ;  and  finally  we  examine  whether  any  or 
all  of  the  values  we  have  found  violate  any  conditions 


/^  OF  THE 

I    UNIVERSITY 


280  PROBLEMS. 

which  are  expressed  or  implied  in  the  problem,  but  which 
are  not  contained  in  the  equations. 

The  following  are  examples  of  problems  which  lead  to 
quadratic  equations. 

Ex.  1.  How  many  children  are  there  in  a  family,  when  eleven 
times  the  number  is  greater  by  live  than  twice  the  square  of  the 
number  ? 

Let  X  be  the  number  of  children  ;  then  we  have 

11  X  =  2x2 +  5; 
.-.  2x2-11x4-5  =  0, 
that  is  (2x-l)(x-6)  =  0. 

Hence  x  =  5,  or  x  =  |. 

Thus  there  are  5  children,  the  value  |  being  inadmissible. 

Ex.  2.  Eleven  times  the  number  of  yards  in  the  length  of  a 
rod  is  greater  by  five  than  twice  the  square  of  the  number  of 
yards.     How  long  is  the  rod  ? 

This  leads  to  the  same  equation  as  before,  only  in  this  case  we 
cannot  reject  the  fractional  result.  Thus  the  rod  may  be  five 
yards  long,  or  it  may  be  half  a  yard  long. 

Ex.  3.  A  number  of  two  digits  is  equal  to  twice  the  product 
of  the  digits,  and  the  digit  in  the  ten's  place  is  less  by  3  than  the 
digit  in  the  unit's  place.     What  is  the  number  ? 

Let  X  be  the  digit  in  the  ten's  place  ;  then  x  +  3  will  be  the  digit 
in  the  unit's  place.     The  number  is  therefore  equal  to 

10  X  +  (x  +  3). 
Hence,  by  the  question, 

10x  +  (x  +  3)=2x(x  +  3)  ; 

.-.  2x2-  5x- 3  =  0, 

that  is  (x  -  3)  (2  X  +  1)  =  0. 

Hence  x  =  3,  or  x  =  —  |. 


PROBLEMS.  281 

Now  the  digits  of  a  number  must  be  positive  integers  ;  hence  the 
second  value  is  inadmissible. 

Therefore  the  digits  are  3  and  6,  and  the  required  number  is  36. 

Ex.  4.  The  square  of  the  number  of  dollars  a  man  possesses 
is  greater  by  1000  than  thirty  times  the  number.  HoW  much  is 
the  man  worth  ? 

Let  X  be  the  number  of  dollars  the  man  is  worth ;  then,  by 
the  conditions  of  the  problem 

a;2  =  30x4-1000; 

.-.  x2-30x- 1000  =  0, 

that  is  (x  -  50)  {x  +  20)  =  0. 

Hence  a;  =  50,  or  x  =  —  20. 

Both  of  these  values  are  admissible,  provided  a  debt  is  considered 
as  a  negative  possession. 

Hence  the  man  may  have  $  50,  or  he  may  owe  $  20. 

Ex.  5.  The  sum  of  a  certain  number  and  its  square  root  is  42 : 
what  is  the  number  ? 

Let  X  be  the  number ;  then 

aj+  ^x  =  42, 

that  is  y/x  =  42  —  X. 

Square  both  sides  ;  then,  after  transposition,  we  have 

x2_85x+ 1764  =  0, 

the  roots  of  which  are  36  and  49. 

The  value  49  will  not  however  satisfy  the  condition  of  the 
problem,  if  by  a  square  root  of  a  number  is  meant  only  the  arith- 
metical square  root. 

Ex.  6.  The  sum  of  the  ages  of  a  father  and  his  son  is  100 
years ;  also  one-tenth  of  the  product  of  their  ages,  in  years, 
exceeds  the  father's  age  by  180.     How  old  are  they  ? 

Let  the  father  be  x  years  old ;  then  the  son  will  be  100  —  x 
years  old.    Hence  by  the  conditions  of  the  problem, 


282  PROBLEMS. 

■^\x(100-x')  =  x+  180; 

.-.  x2-90x  +  1800  =  0, 

that  is  (x  -  60)  (x  -  30)  =  0. 

Hence  x  —  60,  ov  x  =  30. 

The  second  value  is  inadmissible,   although  it  is  a  positive 
integer,  for  it  would  make  the  son  older  than  his  father. 
Hence  the  father  must  be  60,  and  the  son  40  years  old. 

EXAMPLES  LX. 

1.  Find  two  numbers  one  of  which  is  three  times  the  other  and 
whose  product  is  243. 

2.  Find  two  numbers  whose  sum  is  18  and  whose  product  is  77. 

3.  Find 'two  numbers  whose  difference  is  20,  and  the  sum  of 
whose  squares  is  650. 

4.  Divide  25  into  two  parts  whose  product  is  156. 

5.  Divide  80  into  two  parts  the  sum  of  the  squares  of  which  is 
3208. 

6.  A  certain  number  is  subtracted  from  36,  and  the  same 
number  is  also  subtracted  from  30  ;  and  the  product  of  the  re- 
mainders is  891.     What  is  the  number  ? 

7.  A  rectangular  court  is  ten  yards  longer  than  it  is  broad  ;  its 
area  is  1131  square  yards.     What  is  its  length  and  breadth  ? 

8.  The  product  of  the  sum  and  difference  of  a  number  and  its 
reciprocal  is  3| :  find  the  number. 

9.  The  number  of  tennis  balls  which  can  be  bought  for  a  pound 
is  equal  to  the  number  of  shillings  in  the  cost  of  125  of  them. 
How  many  can  be  bought  for  a  pound  ? 

10.  The  number  of  eggs  which  can  be  bought  for  25  cents  is 
equal  to  twice  the  number  of  cents  which  8  eggs  cost.  How 
many  eggs  can  be  bought  for  25  cents  ? 


PROBLEMS.  283 

11.  A  cask  contains  a  certain  number  of  gallons  of  water,  and 
another  cask  contains  half  as  many  gallons  of  wine  ;  six  gallons 
are  drawn  from  each,  and  what  is  drawn  from  the  one  cask  is  put 
into  the  other.  If  the  mixture  in  each  cask  be  now  of  the  same 
strength,  find  the  amount  of  water  and  wine. 

12.  The  cost  of  an  entertainment  was  $  20,  which  was  to  have 
been  divided  equally  among  the  party,  but  four  of  them  leave 
without  paying,  and  the  rest  have  each  to  pay  25  cents  extra  in  con- 
sequence.    Of  how  many  persons  did  the  party  consist  ? 

13.  A  man  buys  a  certain  number  of  articles  for  §  5,  and  makes 
$3.82  by  selling  all  but  two  at  4  cents  each  more  than  they  cost. 
How  many  did  he  buy  ? 

14.  A  man  bought  a  certain  number  of  railway-shares  for 
$9375;  he  sold  all  but  15  of  them  for  $10,450,  gaining  $20  per 
share  on  their  cost  price;  how  many  shares  did  he  buy  ? 

16.  A  crew  can  row  a  certain  course  up  stream  in  8f  minutes, 
and  if  there  were  no  stream  they  could  row  it  in  7  minutes  less 
than  it  takes  them  to  drift  down  the  stream  ;  how  long  would  they 
take  to  row  down  with  the  stream  ? 

16.  A  boat's  crew  can  row  8  miles  an  hour  in  still  water.  What 
is  the  speed  of  a  river's  ciirrent  if  it  take  them  2  hours  and  40 
minutes  to  row  8  miles  up  and  8  miles  down  ? 

17.  Two  trains  run  without  stopping  over  the  same  36  miles  of 
rail.  One  of  them  travels  15  miles  an  hour  faster  than  the  other, 
and  accomplishes  the  distance  in  12  minutes  less.  Find  the  speed 
of  the  two  trains. 

18.  A  person  having  7  miles  to  walk  increases  his  speed  one 
mile  an  hour  after  the  first  mile,  and  is  half  an  hour  less  on  the 
road  than  he  would  have  been  had  he  not  altered  his  rate.  How 
long  did  he  take  ? 

19.  A  and  B  together  can  do  a  piece  of  work  in  a  certain  time. 
If  they  each  did  one  half  of  the  work  separately,  A  would  have 
to  work  one  day  less,  and  B  two  days  more  than  before.  Find  the 
time  in  which  A  and  B  together  do  the  work. 


284  PROBLEMS. 

20.  The  price  of  photographs  is  raised  50  cents  per  dozen  ;  and, 
in  consequence,  six  less  than  before  are  sold  for  $  5.  What  was 
the  original  price  ? 

21.  What  are  eggs  a  dozen  when  two  more  for  30  cents  would 
lower  the  price  2  cents  a  dozen  ? 

22.  A  woman  spends  75  cents  in  eggs  ;  if  she  had  bought  a 
dozen  less  for  the  same  money,  they  would  have  cost  her  3|  cents  a 
dozen  more.     How  many  did  she  buy  ? 

23.  The  price  of  one  kind  of  sugar  per  pound  is  2  cents  more 
than  that  of  a  second  kind,  and  10  pounds  less  of  the  first  kind 
can  be  got  for  $2.40  than  of  the  second.  Find  the  price  of  each 
per  pound. 

24.  One-half  of  the  number  of  cents  which  a  dozen  apples  cost 
is  greater  by  2  than  twice  the  number  of  apples  which  can  be 
bought  for  30  cents.     How  many  can  be  bought  for  $  2.50  ? 

25.  Divide  $  3620  among  A,  B  and  C  so  that  B  shall  receive  $  20 
less  than  A,  and  C  as  many  times  B's  share  as  there  are  dollars 
in  A's  share. 

26.  Find  two  fractions  whose  sum  is  |,  and  whose  difference  is 
equal  to  their  product. 

27.  Two  men  start  at  the  same  time  to  meet  each  other  from 
towns  which  are  25  miles  apart.  One  takes  18  minutes  longer 
than  the  other  to  walk  a  mile,  and  they  meet  in  5  hours.  How 
fast  does  each  walk  ? 

28.  The  men  in  a  regiment  can  be  arranged  in  a  column  twice 
as  deep  as  it  is  broad.  If  the  number  be  diminished  by  206,  the 
men  can  be  arranged  in  a  hollow  square  three  deep,  having  the 
same  number  of  men  in  each  outer  side  of  the  square  as  there 
were  in  the  depth  of  the  column.  How  many  men  were  there  at 
first  in  the  regiment  ? 

29.  The  area  of  a  certain  rectangle  is  equal  to  the  area  of  a 
square  whose  side  is  six  inches  shorter  than  one  of  the  sides  of  the 
rectangle.    K  the  breadth  of  the  rectangle  be  increased  by  one  inch 


MISCELLANEOUS   EXAMPLES   IV.  285 

and  its  length  diminished  by  two  inches,  its  area  would  be  unal- 
tered.    Find  the  lengths  of  its  sides. 

30.  The  diagonal  and  the  longer  side  of  a  rectangle  are  together 
five  times  the  shorter  side,  and  the  longer  side  exceeds  the  shorter 
by  35  yards.     What  is  the  area  of  the  rectangle  ? 

31.  If  the  greatest  side  of  a  rectangle  be  diminished  by  3  yards 
and  the  less  by  1  yard  its  area  would  be  halved  ;  and  if  the  greater 
be  increased  by  9  yards  and  the  less  diminished  by  2  yards  its  area 
would  be  unaltered.    Find  the  sides. 

32.  Two  trains  A  and  B  leave  P  for  Q  at  the  same  time  as  two 
trains  C  and  D  leave  Q  for  P.  A  passes  C  120  miles  from  P,  and 
D*140  miles  from  P.  B  passes  C  126  miles  from  Q,  and  D  half 
way  between  P  and  Q  :  find  the  distance  from  P  to  Q. 

MISCELLANEOUS    EXAMPLES     IV. 
A.  1.  Simplify  2x  -  [Sx  -  9y  -  {2x  -  Sy  -(^x  +  6 y)}]. 

2.  Multiply  a2_|. 25 62+402+6 a6-2ac+ 10 6c  by  a-56+2c. 

3.  Divideic8+(4a6-62)a;_(a_2  6)(a2+3  62)  by  jc-a+2  6. 

4.  Find  the  factors  of  (i.)  (2  x  +  y  -  5?)2  -  (a;  +  2  y  +  4  z)^. 
(ii.)  x'^y^  -x^  -y^+1,  and  (iii.)  x'^y^z^  -  x^z  -  y'^z  +  1. 

(ii.) ^Lul 2         ^-2  '  ^-^ 


(x-2)(x-3)        (x-3)(x-l)      (x-l)(x-2) 
6.   Solve  the  equations : 


X  —  a     x  —  b     X  —  c 
(ii.)  4x2 -25x- 21  =  0. 

7.  Show  that  x2  —  5  x  +  7  can  never  be  less  than  |. 

8.  The  difference  of  the  cubes  of  two  consecutive  integer  num- 
bers is  919  :  find  the  numbers. 


286  MISCELLANEOUS    EXAMPLES   IV. 

B.  1.   Simplify  {x  +  3)3  -  3(a;  +  2)3  +  Z{x  +  1)3  -  x\ 

2.  Show  that 

{x  -  ay  +  (y  -  6)2  +  (a2  +  52  _  Y)^,^  ^y2_y^ 
=  (xa  +  by  -  1)2  +  (6a;  -  ayy. 

3.  Divide  x^  -2  a^x^  -\- a^  by  x^-2ax-\-  a^. 

4.  FindtheL.  CM.  of 

8a;3  +  27,    16x4  +  36x2  +  81,  and  6x2 -5x- 6. 
6.    Simplify 

2  1  x  +  3 


.(i.) 


1        X  +  1        X2  +  1 


Cii  >)  {(q  +  6)(a  +  6  +  c)+  c^Ka  +  hy  -  c^} 
^  '^  {(a  +  6)3  -  c3}(a  +  6  +  c) 

6.  Solve  the  equations : 

(i.)  (6-x)(l  +  2x)+3x(x  +  6)  =  (x  +  l)2-x. 
(ii.)  5x2  +  7x  =  160. 
(iii.)  x2  +  xy  =  10,  y"^  -xy  =  3. 

7.  Show  that  the  sum  of  the  squares  of  the  roots  of  the  equation 
x2-5x  +  2  =  0  is  21. 

8.  At  a  concert  $  6  was  received  for  reserved  seats,  and  the  same 
sum  for  unreserved  seats.  A  reserved  seat  cost  50  cents  more  than 
an  unreserved  seat,  but  6  more  tickets  for  unreserved  than  for 
reserved  seats  were  sold.     How  many  tickets  were  sold  altogether  ? 

C.   1.    Simplify  12a-3{6-2(a-36)-2a}. 

2.  Multiply  a3  +  2  a26  -  a62  +  2  63  by  a3  _  2  a%  -ah'^-2  63. 

3.  Divide  24  x*  -  10  x^y  -  8  x2?/2  +  10  xy^  -  4  2/*  by  2  ?/2-xy-4x2. 

4.  Find  the  factors  of        9  x2  +  9  x  +  2, 

and  of  4(a6  -  cdy  -  {a?-  +  62  -&-  d^y. 


MISCELLANEOUS   EXAMPLES  IV.  287 

6.   SimpUfy  __l  +  xy_^ 

1    <y  -  ^) 

l  +  xy 

and  show  that  ^-::i^  +  ^^1^  +  ^^^  +  ^^  -  ^^^"  -  ^^^  -  ^^  =  0. 
a  6  c  a6c 

6.    Solve  the  equations  : 


3  6  6 

(ii.)  ic  +  1/  =  2  a,  x2  +  ?/2  =  2  a2. 

7.  Find  the  least  value  of  x^  -\-Qx-\-  12,  and  the  greatest  value 
of  6ic-x2-4. 

8.  A  and  B  have  45  coins  between  them,  which  are  all  dollars 
and  dimes.  A  has  four  times  as  many  dollars  as  dimes,  and  B  has 
just  as  many  dollars  as  dimes ;  also  A  has  $  9.50  more  than  B. 
How  much  money  has  each  ? 

D.    1.   Show  that 
(6  +  c)2  _  a2  +  (c  +  ay  -  h"^  +{a  +  by  -  c"^  =  {a  +  h  +  cy. 

2.  Arrange  (1  +  xy  +  2(1  —x-^-x"^)  according  to  ascending 
powers  of  X. 

3.  Show  that  the  difference  between  the  squares  of  any  two  con- 
secutive numbers  is  one  more  than  double  the  smaller  number. 

4.  Show  that  if  -  -|-  -  +  i  =  0,  a*  +  62  +  c2  =(«  +  ft  +  cy. 

a     b      c 

6.  Find  the  L.  C.  M.  of 

a;2  -  5a:  -  14,  x2  -  4a;  -  21,  and  x^-Sx"^-  25x  -  21. 
For  what  value  of  x  will  all  three  expressions  vanish  ? 

6.    Solve  the  equations 

(i.)  4_3^5^6^3^10, 
X     y  X     y 


X     a     b     X—  a  —  b 
(ill.)  x  +  y  =  3,  3  x2  -  11  y2  =:  1. 


288  MISCELLANEOUS   EXAMPLES   IV. 

7.  If  oji,  CC2  are  the  roots  of  ax^  +  bx  +  c  =  0,  prove  that 

Xi,X2_  b^  —  2  ac 
X2     Xi  ac 

8.  Out  of  a  cask  containing  60  gallons  of  alcohol  a  certain 
quantity  is  drawn  off  and  replaced  by  water.  Of  the  mixture  a 
second  quantity,  14  gallons  more  than  the  first,  is  drawn  off  and 
replaced  by  water.  The  cask  then  contains  as  much  water  as 
alcohol.     How  much  was  drawn  off  the  first  time  ? 

E.   1.    Find  the  numerical  value  of 

2.  Show  that  (a  +  by  -  (a^  -  b^y  =  4ab(a  +  by,  and  that 
2(a  -  6)(a  -  c)+  2(&  -  c)(6  -  a)+2(c  -  a)(c  -  b) 

=  {b~  cf  +  (c  -  ay  +  (a  -  5)2. 

3.  Divide  («  +  2  6  -  3  c  +  d)2  -  (2  a-f  6  +  3  c  -  df)2  by  a  +  6. 

4.  Find  the  L.  C.U.oiQx'^  -  bax -Q a^  and  4 a;^  _  2 ax2  -  9 a^. 

6.    Simplify  (1^  +  ^)^ + 
\x       yly'i-x'^ 

6.  Solve  the  equations : 

X  —  b     X  —  a 
(ii.)  {X  -  l)(x  -  2)  +  (a;  -  2)(x  -  3)  +  («  -  3)(x.-  1)=  11. 

(iii.)x  +  l  =  fl,  y+^  =  \ 
?/     10  X     3 

7.  For  what  values  of  x  are  -  +  -  and  -  +  -  equal  to  one  an- 

a     X         a     c 

other  ? 

8.  A  tricyclist  rode  180  miles  at  a  uniform  rate.  If  he  had 
ridden  3  miles  an  hour  slower  than  he  did,  it  would  have  taken 
him  3  hours  longer.     How  many  miles  an  hour  did  he  ride  ? 


Xy  —  y2        a;2  4.  y'l 


MISCELLANEOUS  EQUATIONS.  289 

F.  1.   Add    together    S  a^  -  A  a'^b -\- 2  ah\    Sa'^b-i  ab^  +  2  63, 
3a6-2_4  63  +  2a3,  and  3 63  _  4 a3  4.  2 a26. 

2.  Show  that  (a^  +  ft^  +  c^)  (x^  +  y2  +  z^)  -  (ao;  +  6y  +  czy 

=  {ay  -  6x)2  +  (6z  -  cyy  +  (ex  -  azy. 

3.  Divide  2  +  11  x  +  11  a;2  +  x^  -  x*  by  x2  +  3x  +  2^ 

4.  Find  the  H.  C.  F.  of  x3-x2-2x+2  and  x*-3x3+2x2+x-l. 
What  value  of  x  will  make  both  expressions  vanish  ? 

6.  Simplify    (i.)         ^         ■  1.3 


(ii.) 


x(x-2)      x2-6x  +  6     x(3-x) 
1 


1- 


1-     1 


1-x 

6.  Solve  the  equations : 

(i^  a;4-3     6-x^         7, 
^•^5  10  10 

(ii.)  ^±1  +  EZL|  =  2,  ax  +  6y  =  a2  +  62. 
a  +  6     a—  6 

(iii.)^±i  +  ^±2=:2.^  +  3 


x-1      x-2        x-3 

7.  Find  the  difference  of  the  squares  of  the  roots  of  the  equation 

x2  -  7  X  +  9  =  0. 

8.  What  number  exceeds  its  square  root  by  166  ? 

MISCELLANEOUS   EQUATIONS. 

1.  (x  -  1)  (x  -  4)  +  (X  -  3)  (X  -  5)  =  2(x  -  3)2. 

2.  i(x-2)-K^-4)+K^-5)=0. 


8. 

X  - 

1 

-3      x-5 

X 

1 
-2 

1 
x-4 

4. 

15: 

T 

r  +  17y  = 

79, 

17  X 

+  16y- 

:81, 

290  MISCELLANEOUS   EQUATIONS. 

5.  A  man  drives  to  a  certain  place  at  the  rate  of  8  miles  an 
hour.  Returning  by  a  road  3  miles  longer  at  the  rate  of  9  miles 
an  hour,  he  takes  7|  minutes  longer  than  in  going.  How  long  is 
each  road  ? 


6.  i(x-2)-^^^+5?-=^=0. 


7.  S(x  +  l)(a;  -  3)  +  (x  +  7)(3x  -  13)  =  (2x  -  5)(3x  -  3). 


g    2a;  +  9     3a;-2_5a;  + 14 

'    x  +  2        x  +  3         x  +  4:  ' 

9.  ?  +  y  =  82,  ^  +  y  =  8S. 
7      8  '  8      7 


10.  A  pound  of  tea  and  5  pounds  of  sugar  cost  90  cents,  but  if 
the  tea  were  to  rise  20  per  cent  and  the  sugar  25  per  cent  in  price, 
they  would  cost  f  1.10.     What  was  the  price  of  the  tea  ? 


11.  ^(6x  +  3)+3^(7a;  +  6)+TV(9a^  +  2)=2x+l. 

12.  ?  +  ^  =  «_:^. 
abba 

13   a;  +  3     x  —  6  __x  i- 4:     x  —  5 
' x+1     x-4     x+2     x-S 

14.  X  +  2/  =  2,  (a  +  b)x  +(b-a)y  =  2  b. 

15.  A  poultry  keeper  obtained  100  eggs  more  in  May  than  in 
April,  and  the  daily  average  in  May  was  3  more  than  in  April. 
How  many  did  he  have  in  May  ? 


16.  3a;  -  3[4x  -  2(2x  -  5)]=  9  -  2[3x  -  b{x  -  6)]. 

17.  ^-6  =  2,  4  +  12  =  3. 


X     y  X      y 

18.  15x2  +  34^  +  15  =  0. 

19.  a?  +  2     a;-3_.^^ 
*  X  — 4     »  — 6 


MISCELLANEOUS   EQUATIONS.  291 

20.  A  father's  age  is  three  times  that  of  his  youngest  son,  and 
in  7  years  he  will  be  twice  as  old  as  his  oldest  son  ;  also  the  oldest 
son  is  5  years  older  than  the  youngest.     What  are  their  present 


21.  K^  +  l)(x  +  3)=  i(x  +  5)(x+2)+lix-l)(x^  4). 

22.  (6  +  c)(x-a)-(^c  +  a) (x  -  6)  =  (a  +  b) (x  -  c). 

23.  17x- 19y  =  4,  27x-29y  =24. 
24  2x-3     3x+8_. 

25.  Two  passengers,  who  have  altogether  360  lb.  of  luggage,  are 
charged  80  cts.  for  excess  above  the  weight  allowed  free ;  if  one 
passenger  had  taken  the  same  weight  of  luggage,  he  would  have  been 
charged  $  1.30.     What  weight  is  allowed  free  ? 


26    ^^-  1  J.  5x4-1  _  9;x  +  1  _  1  -X 
5      "^       6  8  3     * 

27.  ^-±l=2x-y-2,  ^y-^  =  y-x  +  5. 

3  2 

28.  (a  +  b)x  +  (a  -  b)y  =  a^  -  6^ 

{a  -  b)x -\- la  +  b)y  =  a^  +  bK 

29.  _i_+_§_=3. 
3-x     8-x 

30.  A  father's  age  is  equal  to  the  united  ages  of  his  five  children, 
and  five  years  ago  his  age  was  double  their  united  ages.  How  old 
is  the  father  ? 


31      2x       x  +  3  _  3x+  19 
'x-lx+1       x+6  • 

32.  l{x-2)=l(il-y),  26x  +  32/  +  4=0. 

33.  1  17         1 


x2-l     1-a;     8     x  +  1 


34.    y/x  +  Vbx  +  l=2. 


292  MISCELLANEOUS   EQUATIONS. 

35.  Two  cyclists  started  at  the  same  time,  and  one  rode  from 
B  to  A  and  the  other  from  A  to  B  along  the  same  road.  They 
reached  their  destinations  3  hours  and  1  hour  20  minutes,  re- 
spectively, after  they  passed  each  other.     When  did  they  meet  ? 


39.  \/4  X  +  4  -  Vx-\-S  =  Vx  -  4. 

40.  A  certain  sum  of  money  was  divided  between  A,  B,  and  C. 
A  had  one-ninth  of  the  shares  of  B  and  C  together,  B  had  one- 
third  of  the  shares  of  C  and  A  together,  and  C  had  6  dollars  more 
than  A  and  B  together.     How  much  had  each  ? 


"•  i(-!)-(^-T)--«- 

42.  ^±^-^=^  =  1. 
x  —  S     X  —  4 


43.  x-\-2=y/{4  +  xVS-x}. 

44.  3  X  -  2  y  =  12,   9  x2  -  4  2/2  =  576. 

45.  A  man  about  to  invest  in  the  2f  per  cents  observed  that,  if 
the  price  of  the  stock  had  been  7|  less,  he  would  have  got  f  per 
cent  more  interest  for  his  money.  What  was  the  price  of  the 
stock  ?  

j,^    4x^  +  4x2  +  8x4-l_2x2  +  2x+l. 

x  +  1 

5)  =11,   3,V(55y-12)  =  ^-37. 


2  x2  +  2  X  +  3 

47. 

tV(5^  +  6)-^V(112/ 

48. 

X        X—  1 

49. 

V«  +  Vx  +  4  =  -^' 

MISCELLANEOUS   EQUATIONS.  293 

50.  A  fast  train  left  Cambridge  for  London  at  the  same  time  as 
a  slow  train  left  London  for  Cambridge.  They  arrived  at  their 
destinations  three-quarters  of  an  hour  and  one  hour  and  twenty- 
minutes  respectively  after  they  passed  one  another.  Find  how 
long  each  took  for  the  journey. 


x-1     SVx-l      3J 


23 


lO(x-l) 

52.  6x  +  2y  -l=Sx-y  +  U  =  x-^19y  +  6. 

53.  V(ic2-a2)=2-  ^ 


y/ix^-a^) 

54.  3x2  -  2 a;?/  +  4 1/2  =  36,   4x'^-y^  =  7. 

55.  Find  three  consecutive  integers  such  that  the.  sum  of  the 
cubes  of  the  greatest  and  least  exceeds  twice  the  cube  of  the 
middle  number  by  42. 


66    ^--^     ^ _2a;-l     9}  -  x 

3  6      ~       3      "•"     21    * 

57    a;-3     x-2         2      .       3 


2  3         x-S     x-2 

58.  ax  -  &?/  =  a2  -  62  _  2  ab,   bx -\-  ay  =  2  ab -\- a^  -  b\ 

59.  a;2  4-  a;?/  +  y2  _  g^  ^^  4.  y^^yi  +  y*  =  243. 

60.  A  man  bought  a  certain  number  of  articles,  of  which  he  sold 
half  at  5  per  cent  profit,  one-third  at  10  per  cent  profit,  and  the 
rest  at  30  cents  each.  His  total  profit  was  at  the  rate  of  If  per 
cent.    What  was  the  cost  of  each  article  ? 


61.    (a-f  b)(x-\-a-b)  +  {a-  6)(x  -  a  -  6)+ 2a(a;  -  2c)=0. 

62    3a;-2     2a;-3^2 
2x-b     3x-2     3* 

63.  2x2 -3V(«2  +  2x-M4)-f4x- 49  =  0. 

64.  2x2+ 3x2/ =  8,   y'^-2xy  =  29. 


294  MISCELLANEOUS   EQUATIONS. 

65.  A  man  bought  a  certain  number  of  sheep  for  $  290.  Having 
lost  five  of  them,  he  sold  one  quarter  of  the  remainder  for  $  63, 
making  a  profit  of  6  per  cent  on  the  sale  of  these.  How  many 
did  he  buy  ? 

66.  1  :.  +  5     _      3 


6a; -1      1-25x2      1  +  5x 

67.  (a'+  h)x  +  hy-ax^-{a+  b)y  =  a^  +  &». 

68.  ^(x-a)  +  y/(x-b):  ^  '  ^ 


y/(x-a)      ^(x-b) 

69.  x"^-^  y^  =  xy  +  7,   x^  —  y^  =  xy  -  1. 

70.  A  number  of  articles  were  bought  for  $  25,  and  sold  again 
at  an  advance  of  12 1  cents  in  the  price  of  each  ;  and  at  the  advanced 
price  .1 25  was  received  for  the  sale  of  all  but  10  of  the  articles. 
How  many  were  there  ? 


71.  \iSx  +  1)~  ^(2x  -  S)=  j'^{6x  -  1)-  i(7 X  -3). 

72.  ---  —  ---. 
c      x~  a'     c 

73.  ^(x+l)  +  V2^=^(i6x  +  l). 

74.  x^  -xy-\-Sy  =  ll,   y^  -  xy  -Sx  +  1  =0. 

75.  The  value  of  185  coins  consisting  of  dollars,  dimes,  and 
half-dimes  amounts  to  $30.50.  If  there  were  twice  as  many 
dimes,  half  as  many  half-dimes,  and  three  times  as  many  dollars, 
the  total  value  of  the  coins  would  be  $  72.  How  many  coins  are 
there  of  each  kind  ? 


-g    3x+l      l-5x_ic-fl      2-9« 


5                3              3 

77. 

x+1      a;2  4.  1  _ 29 
x^  +  lz  +  1      10 

78. 

(x  -a)(y  -b)  =  ab,   bx  =  ay. 

79. 

x2  =  2y-f24,   ?/2  =  2x  +  24. 

MISCELLANEOUS   EQUATIONS.  295 

80.  1120  square  feet  of  paper  will  just  cover  the  four  walls  of  a 
room  which  is  8  feet  longer  than  it  is  wide ;  but,  if  the  room  were 
4  feet  higher,  the  same  quantity  of  paper  would  just  cover  the  two 
smaller  and  one  of  the  larger  walls.  What  are  the  dimensions  of 
the  room  ? 

81.  Ka^_2)-^  +  5^  =  0. 

82.  (a-xy+{b-xy=ia-\-b-2x)^. 

83.  (x-6)2+(y-6)2  =  2(a;y-40),   x  =  y  +  l. 

84.  4.xy  =  96-x^y%  x  +  y  =  6. 

85.  A  merchant  gained  as  many  eagles  on  the  sale  of  a  certain 
quantity  of  coal  as  there  were  half-dollars  in  the  cost  price  of  a 
ton,  or  half-dimes  in  the  retail  price  of  a  cwt.  How  many  tons 
did  he  sell  ? 


86    2x-3     2x-4_2x-6     2a;-7 
*  2ic-4     2a;-6~2a;-7     2a;-8* 
87.   2xCa-b-2x)  +  ia-l)(b-l)=0. 


88.  y/(6x+l)  +  V2{l-x)=zy/7x-\-6. 

89.  xy(x  +  y)  =  30,  x^-\-y»  =  -  91. 

90.  A  rectangular  enclosure  is  half  an  acre  in  area,  and  its 
perimeter  is  198  yards.    Find  the  length  of  its  sides. 


91      g;  +  3       3a;-|-7_2a;  + 1 


2a;+l     2a;-2      x-2 

92.  x-\-2y  -\-Zz  =  i,  x  +  Sy  +  2  =  iz,  x -\- 2z -\- S  z=  iy. 

93.  ^ia-x)  +  -^(x-b)  =  ^(ia-b). 

94.  {x  -  3)2  +(2/  -  3)2  =  34,  xy  -  3(x  +  y)=  6. 

95.  Two  trains  start  at  the  same  instant,  the  one  from  B  to'  A, 
the  other  from  A  to  B,  the  distance  between  A  and  B  being  100 
miles.  The  trains  meet  in  1  hour  15  minutes,  and  one  train  gets 
to  its  destination  1  hour  20  minutes  before  the  other.  Find  the 
rates  of  the  trains. 


296  MISCELLANEOUS   EQUATIONS. 

gg    ax  -  hy  _{2  a  +  h)x  -  (^a  +  2  b)y  _  ^^     ^^ 
3  7 

97.  13x2+ 12  =  80  ic. 

98.  (a:  +  l)(x  +  2)(x  +  4)(x  +  6)  =  4. 

99.  x^iy  +  l)+y^{x  +  l)=  109,  xy  =  12. 

100.  A  starts  to  walk  from  P  to  Q  at  10  a.m.,  B  starts  from  Q  to 
walk  to  P  at  10.24  a.m.  They  meet  6  miles  from  Q.  B  stops  1 
hour  at  P,  and  A  stops  2  hrs.  54  min.  at  Q,  and  returning  they 
meet  midway  between  P  and  Q  at  6.54  p.m.  Find  the  distance 
from  P  to  Q. 


101.  y^  +  z'^  =  z^  +  x'^  =  x^-^y^  =  axyz. 

102.  ax(y  -\-  z)=  hy{z  -{■  x')  =  cz(x  +  y)—  xyz. 

103.  x'^  ^-  2yz  =  y"^  -\-  2.  zx  =  z"^  -^  2xy  =  12. 

104.  Resolve  x^  -\-  y^  -\-  z^  —  Z  xyz  into  the  three  factors 

X  -\-  y  -\-  z,  X  +  cjy  +  a>%,   X  +  <a^y  +  uz, 

where  w  =  —  ^  +  ^V^, 

and  determine  three  alternative  relations  between  a,  6,  and  c,  one 
of  which  must  be  satisfied  in  order  that  the  equations 

x^  +  y^  -^  z^  —  S  xyz  —  0, 

ax+  by  +  cz  =  0, 

a     b     c 
may  be  simultaneous  in  x,  y,  and  z. 


POWERS  AND  ROOTS.  297 


CHAPTER  XX. 

Powers  and  Roots. 

196.  The  process  by  wMcli  the  powers  of  quantities 
are  obtained  is  called  involution ;  and  the  inverse  process, 
by  which  the  roots  of  quantities  are  obtained,  is  called 
evolution.     [Art.  9.] 

INVOLUTION. 

196.  When  m  and  n  are  any  positive  integers,  we  have 
by  definition 

a*"  =  aaaa. . .  to  m  factors, 

and       a"  =  aaaa...  to  n  factors  ; 

.  • .  a^xa*'  =  (aaaa. . .  to  m  factors)  x  (aaaa. . .  to  n  factors) 

=  aaaa...  to  m  +  w  factors 

=  a"*+**,  by  definition. 

Hence  when  m  and  n  are  any  positive  integers, 

^m  _|_  ^n  _  ^m+n 

Thus,  the  index  of  the  product  of  any  two  powers  of  the 
same   quantity  is  the  sum  of  the  indices  of  the  factors. 
This  result  is  called  the  Index  Law. 
From  the  Index  Law  we  have 

a"^  xa""  xa^  =  »"*+"  X  a"  =  a'"+"+^, 

and  so  on,  however  many  factors  there  may  be. 


298  POWERS   AND   ROOTS. 

Hence  a"*  x  a**  x  a/  •  •  •  =  a'^+^+p+-. 

Thus,  the  index  of  the  product  of  any  number  of  powers 
of  the  same  quantity  is  the  sum  of  the  indices  of  the  factors. 

197.   We  have 

ttxaxaxa...tom  factors 


a'^/a''  = 


axaxax  a...  to  n  factors 


Now,  if  m  is  greater  than  n,  the  n  factors  of  the  denom- 
inator can  be  cancelled  with  n  of  the  factors  of  the 
numerator :  we  then  have  m  —  n  factors  left  in  the 
numerator. 

Thus,  when  m  is  greater  than  w, 

If,  however,  n  is  greater  than  m,  the  m  factors  of  the 
numerator  can  be  cancelled  with  m  of  the  factors  of 
the  denominator :  we  then  have  n  —  m  factors  left  in 
the  denominator. 

Thus,  when  m  is  less  than  n, 

198.   To  find  (a'")"  when  m  and  n  are  positive  integers. 
By  definition 

(oC^Y  =  a*"  x  a*"  x  a"*  X  •••  to  71  factors 

Hence  («'")'*  =  a"*^ 


POWERS   AND   ROOTS.  299 

Thus,  to  raise  any  power  of  a  quantity  to  any  other 
power,  its  original  index  must  he  multiplied  by  the  index 
of  the  power  to  which  it  is  to  be  raised. 

199.  To  find  (ab)^. 

{ab)'^=  ab  X  ab  X  ab...  to  m  factors,       by  definition, 

=  {aaa. . .  to  m  factors)  x  (bbb...  to  m  factors)     [Art.  52.] 

=  a"*  X  6"",  by  defijiition. 

Hence  (ab)"*  =  a'"6'*. 

Similarly 

(abc)"^  =  abc  x  abc  x  abc...  to  m  factors, 

=  (aaa...  to  m  factors)  x  (bbb...  to  m  factors) 

X  (ccc...  to  m  factors) 
=  a"*  xb"^  X  C*. 

Hence  (abc)"^  =  a'*6"*c*, 

and  so  on,  however  many  factors  there  may  be  in  the  ex- 
pression whose  mth  power  is  required. 

Thus,  the  mth  power  of  a  product  is  the  product  of  the 
rath  powers  of  its  factors. 

200.  The  most  general  monomial  expression  is  of  the 
form  a'b^c'... 

To  find  (a'b"c'...y. 

(a'b*C...y  =  {a''y{b«y{c'y...     [Art.  199.] 
=  a'^b^'^c'"*...  [Art.  198.] 

Thus,  any  power  of  a  monomial  expression  is  obtained  by 
taking  each  of  its  factors  to  a  power  whose  index  is  the 
product  of  its  original  index  and  the  index  of  the  power  to 
which  the  whole  expression  is  to  be  raised. 


300  POWERS   AND   ROOTS. 

201.   The  following  is  an  important  case. 


To  find  '^ 


-)  =-x-x---«tom  factors,  by  definition, 
oj        0       0       0 

a  X  a  X  a...  to  m  factors         p  .   ,    1  fi7  1 
b  X  b  X  b...  to  m  factors 

^. 

b)  ~b^' 

202.  It  should  be  noticed  that  all  powers  of  a  positive 
quantity  are  positive,  and  that  successive  powers  of  a 
negative  quantity  are  alternately  positive  and  negative. 
This  follows  at  once  from  the  Law  of  Signs ;  for  we  have 

(_a)2  =  (-a)(-a)  =  +  a^ 

(-ay  =  {-ay{-a)  =  i-\-a'){-a)  =  -a'; 

(-ay  =  {-ay{-a)  =  {-a')(-a)  =  +  a'', 

and  so  on. 

Thus        (  _  a)  2"  =  -f  a^%  and  (  -  a)  2"+i  =  -  a^^+i. 

From  the  above  it  is  clear  that  all  even  powers,  whether 
of  positive  or  of  negative  quantities,  are  positive,  and  that 
all  odd  powers  of  any  quantity  have  the  same  sign  as  the 
original  quantity. 

203.  We  have  already  proved  the  following  cases  of 
the  involution  of  binomial  expressions. 


POWERS   AND   ROOTS.  301 

and  (a  +  by  =  a'-hSa'b  +  3ab'  +  ¥. 

If  we  multiply  again  hj  a-\-b,  we  shall  have 

(a  +  by  =  a*  +  4a36  +  Ba^ft^  +  4.ab^  +  6*. 

By  multiplying  the  last  result  by  a  +  6  we  should  obtain 
(a-^by,  and  by  continuing  the  process  we  could  obtain 
any  required  power  of  a  +  6 ;  but  to  find  in  this  way  any 
high  power,  for  instance  to  find  (a  +  b)^,  would  clearly 
be  very  laborious. 

We  shall  shortly  prove  a  theorem,  called  the  Binomial 
Theorem,  which  will  enable  us  to  write  down  at  once  any 
power  of  a  binomial  expression. 

The  above  formulae  are  identities,  and  are  true  for  all 
values  of  a  and  b ;  hence  we  can  write  down  the  squares 
and  the  cubes  of  any  binomial  expressions. 

Thus       (a*  -  ¥y  =(a4)2+  2(a4)(-  6*)  +  (-  b^y 
=  a8-2a*6*  +  6S 
and         (5  a2  -  3  62)3  =  (5 ^2)8  +  3(5  o2)2( _  3  52) 

+  3(5a2)(-3  62)2^.(_3ft2)8 

=  125  a»  -  225  a*b^  +  135  a^b*  -  27  6«. 
Also    (a  +  6  +  cy={a  +  (6  +  c)}« 

=  a8  +  3a2(6  +  c)+  3a(6  +  cy+{b  +  cy. 

204.  An  important  case  of  involution  is  considered  in 
Art.  70,  where  the  square  of  a  multimonial  expression  is 
obtained. 


302  POWERS   AND   EOOTS. 

EXAMPLES  LXI. 
Write  down  the  value  of  each  of  the  following : 


1. 

(a3)5.  - 

21. 

(«3>  63)8. 

2. 

(X5)3. 

22. 

(2  a2- 3  62)3. 

3. 

(-a2)8. 

23. 

(3a2_2  62)8. 

4. 

(-  a3)2. 

24. 

(a2  +  62  +  c2)2. 

6. 

i-2a^y. 

25. 

(a3-2  63  +  3c8)2. 

6. 

C-sa^y. 

26. 

(02  _  4  ^,2  _  3  c2)2. 

7. 

i-ab^y. 

27. 

(x2_3x-6)2. 

8. 

{a^h^y. 

28. 

(3^2 -X-  5)2. 

9. 

i-ab^y. 

29. 

(2x2  +  5x-l)2. 

10. 

i-Sa'^b^cy. 

30. 

(3x2 -6x- 6)2. 

11. 

C-ab'^c^y. 

31. 

(1+X  +  X'^  +  X8)2. 

12. 

(-5a263c4)8. 

32. 

(X3  -  X2  +  X  -  1)2. 

13. 

(-Y- 

33. 

(X3  +  x2  -  2  X  -  2)2. 

Wl 

34. 

(a  +  26  +  3c  +  4d)2. 

14. 
16. 

[     bH^j 

35. 
36. 
37. 

(2a-6  +  c-2c?)2. 
(x2  +  X  +  1)8. 

(X2  -  X  +  2)8. 

16. 

(2  a*  +  3  63)2. 

38. 

(3x2-6x  +  l)8. 

17. 
18. 

(a^  -  2  6^)2. 

39. 

l-*lh\{'-w 

19. 
20. 

(-a2  +  2a6)2. 

(a2  +  62)8. 

40. 

'-^^4- 

THE  BINOMIAL  THEOREM. 


205.  The  numerical  coefficients  on  the  right  side  of 
the  formulae  of  Art.  203  may  be  so  constructed  as  to 
exhibit  the  law  of  their  formation.    Thus,"  written  in  the 


POWERS  AND   ROOTS.  303 

order  of  their  occurrence  in  the  formulae,  they  and  their 
reconstructed  equivalents  are 

1,   2,    1  =  1,   |,    |l|,  v|P^ 

1,    3,    S,    1-1,   -,    j-^,    ^-^-^, 

1     A    fi    i    1-1     f    4-3    4-3.2    4-3.2.1 
1,    4,    b,    4,    1-1,    ^,    ^    ^,    ^-^Tg,    1.2.3.4- 

If  this  law  of  their  formation  obtains  for  higher 
powers  of  (a  4-6),  we  shall  be  able  to  write  out,  by  a 
very  simple  rule,  the  entire  series  of  terms  in  the  expan- 
sion of  (a  +  by,  wherein  n  is  any  positive  integer.  Thus, 
if  the  law  holds  for  all  positive  integral  values  of  n, 

{a  +  by  =  a'  -h  5a^6  +  ^a'^'  + 1^4^«'&' 

and  in  general 

(a  +  by  =  a"  +  na^-'b  +  ^(^--^)a"-^>  +  ... 

J.  *  ^ 

1.2.3...r  -- 

Here  r  is  some  integer  less  than  w,  and 

n(n-l)...  (n-r  +  l)^n-.y 
1.2.3...r  ' 

which  is  called  the  general  term,  may  be  made  to  assume 
in  succession  the  form  and  value  of  every  term  in  the 
series,  by  giving  to  r  the  successive  values 
0,  1,  2,  3,  4,  ...n-1,  n. 


304  POWERS   AND   ROOTS. 

Thus,  if  r  =  5,  this  general  term  becomes 

n{n-l)(n-2)(n-S)(n-4:)       ,, 
1.2.3.4.5 

We  proceed  to  prove  that  the  above  formula  is  true 
for  all  positive  integral  values  of  n.  The  proof  here 
given  is  by  mathematical  induction  (monomorphic  trans- 
formation) explained  in  Art.  145. 

For  brevity's  sake  we  write  the  formula  thus : 

(a  +  6)"  =  Coa'*  -f  Cia'-^d  +  C2a''-"-b^  H 

in  which  the  letters  c  have  the  respective  values 

-,  71(71  — 1) 

Co  =  l,  Ci=n,  c2  =  -5^— y^,  •.• 

V 

r  _y^(^-l)  •••  (n-r+1)  _. 

"'"  1.2.3...r  '      ^'''~ 

We   assume    provisionally   that  the    formula  is   true 
when  the  index  is  n.     Can  we  justify  this  assumption  ? 
Multiplying  both  sides  of  it  by  (a  +  6),  we  obtain 

(a  +  6)"+^  =  Coa"+^  +  (co  +  c{)  a^^h  +  (ci  +  Cg)  a'^-^t^ 

+  (C2  +  C3)  a"-263  +  ^.  +  (c,_i  4-  c,)  a"-'-+i6''+  ... 

But       Co  +  Ci  =  n  +  1, 

c,  +  c.  =  .  +  -i^  =  ^^(^, 

'       '         1.2  1.2.3 

_yi(n  — l)(ii  +  l) 
1.2.3 


POWERS  AND  ROOTS.  305 

and  in  general 

_n(n—iy*'(n—r+2)      n(n  — 1)  »••  (n— r+ 1) 
Cr-i-\-Or-       i.2.3...^_i       +  1.2.3...r 

_n(n  —  l)  '"  (n  —r  -\-2)(r  -{-n  —  r  -\'l) 
"■  1.2.3-..r 

_  (yi  +  l)n(n  — 1)  ...  (n +  1  —  ^  +  1) 
1.2.3...r 

Substituting  these  results  for  Cq,  Cq  -\-  Ci,  Cj  +  Cg,  etc.,  in 
the  formula,  we  have 

(a  +  by+^  =  a"+i  +  (w  +  l)a"6  +  ^^"^  ^^^a"-^6^  +  ... 

1  *  ^ 

4.  (n  +  l)n(n-l)...  (n  +  l-r  +  1)    ^^i_, , 
"^  1.2.3...r 

+  •••  +6"+'. 

The  series  thus  arranged  is  exactly  the  same  in  form 
as  that  for  (a  +  6)",  w  +  1  having  now  taken  the  place  of 
n.  Hence,  if  the  formula  be  true  for  any  positive  inte- 
gral value  of  n,  it  is  true  when  n  is  replaced  by  the  next 
higher  integer. 

But  it  is  obviously  true  when  w  =  1 ;  hence  it  is  true 
when  n  =  2.  And,  being  true  when  n  =  2,  it  is  true  when 
n  =  3,  and  so  on  indefinitely.  Therefore  it  is  true  for 
every  positive  integral  value  of  n.     [Art.  145.] 

The  proposition  thus  established  is  known  as  the 
Binomial  Theorem  for  positive  integral  exponents.  It  is 
one  of  the  most  important  theorems  of  algebra,  and  will 
be  discussed  more  fully  in  a  subsequent  chapter  to  be 
devoted  to  this  special  topic. 
u 


306  POWERS   AND   ROOTS. 

The  special  numerical  coefficients  which  have  been 
designated  above  by  Cq,  q,  C2,  c^,  •••  c^,  •••  c„  are  called  the 
binomial  coefBcients.  They  are  frequently  used  in  other 
algebraic  formulae  and  should  be  committed  to  memory. 

Ex.  1.     If  in  the  binomial  formula  we  write  w  =  3,  we  have 

(a  +  6)3  =  a3  +  § ^25  ^hlab^  +  b^ 

=  a3-j.  3^25 +  3^52+  53. 

Ex.  2.    If  a  =  2,  6  =  ?/2^  and  w  =  3,  we  have 

(2 +  y)3  =  23+1-22.2/2  +  ^.22/4 +  2^ 

=  8  +  12^2  +  62/4+2/6. 


Ex.  3.     If  a  =  -,  6  =  -,  and  w  =  4,  we  have 
2  y 


/i  +  iV=l  +  i.^     i  +  ili    1     1   I  ^--^-^    i    A  +  ± 
V2     y)       24      l*23'2/'^1.2*22'2/2     1.2."'-^*    ^^       ^ 


•2.1.1.1 
3*2*2/^2/* 


l  +  l.i  +  §.l+2.1  +  i. 
16     2    2/2    2/^  2/3     y* 


206.  The  formula  for  the  expansion  of  (a  — &)"  is 
obtained  by  writing  —  6  in  place  of  h  in  the  previously 
written  formula 

(a  +  by  =  a'*  +  Cia"-i6  +  i^a'^-^h'^  +  Cga^-^d' H h  6", 

the  even  terms  of  which  become  negative  by  this  change, 
because  odd  powers  of  a  negative  quantity  are  negative. 
Thus, 

(a  -  6) "  =  a"  -  c^al'-^h  +  c^a^'-^h'^  -  c^a^^-^b^  +  •  •  •  ±  &", 

and  the  last  term  has  the  positive  or  the  negative  sign 
according  as  n  is  odd  or  even. 


POWERS   AND   ROOTS.  307 

Ex.  1.  If  a  =  x,  h  =  2y,  and  w  =  4,  the  result  of  substitution 
in  the  formula  for  the  expansion  of  (a  —  &)"  is 

(x  -  2  2/)4  =  X*  -  4  x^(2  y)  +  — x2(2  yy 
1  *  ii 

=  X*  -  8  a;8y  _|.  24  x2«/2  _  32  a;y8  +  16  y*. 

Ex.  2.  If  a  =  2  X,  5  =  3  2/2,  and  n  =  5,  the  result  of  substitu- 
tion in  the  formula  for  the  expansion  of  (a  —  6)"  is 

(2 X  -  3  y2)6  =  (2 xy  -  6(2  x)*(3  y^)  +  ^  (2  x)8(3  y2)2 

-(3  1/2)6 

=  32  x5  -  240  X42/2  +  720  x^j/*  -  1080  x^  +  810  xy8 
-243yw 

207.  General  Term.  By  inspection  of  the  formula,  it  is 
at  once  obvious  that  the  number  of  any  term  in  the 
expansion  of  (a  ±  6)"  is  one  greater  than  the  suffix  of  c 
(or  number  of  factors  in  the  denominator  of  c)  in  that 
term.  Thus  c^a'^'^h^  is  the  fourth  term.  Hence,  in  gen- 
eral, the  (r  4-  l)th  term  is 

n(n-l)(n-2)...(n-r-H) 
"^  1.2.3...r  ' 

and  by  means  of  this  formula  any  isolated  term  may  be 
written  down. 

Observe  that  r  is  at  once  the  exponent  of  h,  the 
number  of  factors  in  the  denominator  and  the  number  of 
factors  in  the  numerator,  and  that  the  sum  of  the  expo- 
nents of  a  and  h  is  n. 


308  POWERS   AND   ROOTS. 

Ex.  1.    The  eleventh  term  in  the  expansion  of  (1  +  x^y^  is 
15.14.13.12.11.10.9.8.7.6 


1.2.3.4.5.6.7.8.9. 10 


(a:2)io  =  3003x20. 


Ex.  2.     The  sixth  term  in  the  expansion  of  (x  —  ^y'^y^  is 
_  11. 10. 9. 8. 7  231 

1.2.3.4.5       ^^^^  16     ^ 

The  sign  of  this  term  is  negative  because  even  terms  in  the  ex- 
pansion of  (a  —  6)"  are  negative. 


EXAMPLES  LXII. 
Write  out  the  following  expansions  : 

1.  (x  +  ay.  3.    (Sx-2yy.  6.    (2x2-1)6. 

2.  (l-x2)5.  4.    (2a  +  3«2)4.  6.    (y-xy. 

7.  Find  the  third  term  of  (a  -  3  &)io. 

8.  Find  the  fifth  term  of  (2x  -  x^y^. 

9.  Find  the  sixth  term  of  (2  a  -  1)8. 

10.  Find  the  seventh  term  of  (1  —  xy^. 

11.  Find  the  eighteenth  term  of  (1  +  x)^. 

12.  Find  the  twenty-first  term  of  (1  -  x)^. 

13.  Expand  (a  +  ^/by  +  (a  -  ^/by. 

14.  Expand  (a  -f  ^by  +  (a  -  ^/by. 

15.  Expand  (a  +  y/by  -f  (a  -  y/by. 

16.  Find  the  middle  term  of  (1  +  xy. 

17.  Find  the  middle  term  of  (1  -f  xy^. 

18.  Find  the  middle  term  of  (2  x  -  3  yy. 

19.  Show  that  in  the  expansion  of  (1  +  a;)'»+'',  the  coefficients 
of  x"*  and  x"  are  equal. 


20.  Expand  (^  +  -)  * 

21.  Expand  ^^x  +  ij-i^x-^)*- 


POWERS   AND   ROOTS.  309 

EVOLUTION. 
We  know  that  there  are  two  square  roots  of  a?, 
namely   ±  a ;  we  also  know   [Art.  188]   that  there  are 
three  cube  roots  of  o?,  of  which  a  is  one,  and  the  other 
two  are  imaginary. 

There  is  therefore  an  important  difference  between 
powers  and  roots ;  for  there  is  only  one  nth  power ^  but 
there  is  more  than  one  nth  root,  of  a  given  expression. 

209.  An  expression  which  when  raised  to  the  ?ith 
power,  where  n  is  any  positive  integer,  becomes  equal  to 
a  given  expression,  is  called  an  nth  root  of  the  given 
expression. 

We  have  shown  in  Art.  199  that  the  mth  power  of  a 
product  is  the  product  of  the  mth  powers  of  its  factors ; 
hence,  conversely,  the  mth  root  of  a  product  is  the 
product  of  the  mth  roots  of  its  factors.* 

Thus  ^ahc  —  ^a  ^h  ^c, 

and  Vah  =  ^a  ^h. 

Again,  we  have  shown  in  Art.  200  that  the  nth  power 
of  a  monomial  expression  is  obtained  by  multiplying  the 
index  of  each  of  its  factors  by  n. 

It  follows  conversely  that  if  we  divide  the  index  of 
each  factor  of  a  given  expression  by  n,  we  shall  obtain 
an  nth  root  of  the  expression.     For  by  raising  to  the  nth 

*  It  should  be  noticed  that  the  proof,  in  Art.  52,  that  the  factors 
of  a  product  may  be  taken  in  any  order,  only  holds  good  when  those 
factors  represent  integral  or  fractional  numbers,  and  does  not  enable 
us  to  assert  that  VaxVh  =  Vhx  Va,  when  Va  or  Vh  is  really  a  surd. 
For  a  proof  that  surds  obey  the  Fundamental  Laws  of  Algebra  see 
Treatise  on  Algebra,  Art.  162. 


310  POWERS   AND   ROOTS. 

power  the  result  obtained  by  such  division  of  the  indices, 
we  must  clearly  get  the  original  expression. 


Thus  one  value  of  -y/a*  is  a^^  one  value  of  Vc^h^^  is  aH>H^^  and 
one  value  of  Va^b^  is  aPb^. 

When  the  square  root  of  an  expression  which  is  not  a 
perfect  square,  or  the  cube  root  of  an  expression  which 
is  not  a  perfect  cube,  is  required,  the  operation  cannot 
be  performed.  We  can,  for  example,  only  write  the 
square  root  of  a  as  ^a,  and  the  cube  root  of  a^  as  -^/a^, 
and  similarly  in  other  cases. 

SQUARE   ROOT. 

210.  We  now  proceed  to  consider  the  square  root  of 
multimonial  expressions. 

In  Art.  114  we  have  shown  how  to  write  down  the 
square  root  of  any  trinomial  expression  which  is  a  com- 
plete square. 

Having  arranged  the  expression  according  to  ascending 
or  descending  powers  of  some  letter,  the  square  root  of 
the  whole  expression  is  then  found  by  taking  the  square 
roots  of  the  extreme  terms  with  the  same  or  with  diffei- 
ent  signs  according  as  the  sign  of  the  middle  term  is 
positive  or  negative. 

Thus,  to  find  the  square  root  of 

4a«-12a463^96«. 
The  square  roots  of  the  extreme  terms  are 

±2a^and  ±36«. 
Hence,  the   middle   term  being  negative,  the  required 
square  root  is  ±  (2  a*  —  3  b^) . 


POWEES   AND   EOOTS.  311 

Note.  — In  future  only  one  of  the  square  roots  of  an  expression 
will  be  given,  namely  that  one  for  which  the  sign  of  the  first  term 
is  positive  :  to  find  the  other  root  all  the  signs  must  be  changed. 

As  other  examples 

V(49  aW  +  28  a5  +  4)  =  7  a5  +  2 
V(l  +  6xy3  +  ^^xV)=  1  +  f  a^y^, 
and  V{«^  +  2  a(6  +  c)  +  (6  +  c)2}=  a -\- b -\- c. 

211.  When  an  expression  which  contains  only  two 
different  powers  of  a  particular  letter  is  arranged  accord- 
ing to  ascending  or  descending  powers  of  the  letter,  it 
will  only  contain  three  terms.  For  example,  the  ex- 
pression 

a^-^b^  +  c^  +  2bc-\-2ca-\-2ab, 

which  contains  no  other  power  of  a  but  a^  and  a,  when 
arranged  according  to  powers  of  a,  is 

a2  +  2a  (6  +  c) -f  (6^  +  c^  +  26c). 
Thus  any  expression  which  only  contains  two  different 
powers  of  a  particular  letter  can  be  written  as  a  trinomial 
expression  ;  and  since  we  can  write  down  the  square  root 
of  any  trimonial  expression  which  is  a  complete  square, 
it  follows  that  the  square  root  of  any  expression  which 
is  a  complete  square  can  be  written  down  by  inspection, 
provided  that  the  expression  only  contains  two  different 
powers  of  some  particular  letter. 

For  example,  to  find  the  square  root  of 

a2  +  62  +  c2  -h  2  6c  +  2  ac  +  2  a6. 

Arranging  the  expression  according  to  the  descending  powers  of 
a,  we  have  a'^  +  2  a{h  +  c)  +  (62  -f  2  6c  +  c2), 

that  is,  a2  +  2  a(&  -1-  c)  +  (6  -H  c)2, 

which  ifl  {a  -h  (6  +  c)f. 


312  POWERS   AND   ROOTS. 

Thus     V(«^  +  62  +  c2  +  2  6c  +  2  ca  +  2  a6)  =  (a  +  6  +  c). 
Also,  to  find  the  square  root  of 

(c*  +  4?/4  4-  9^4  +  4icV  _  6a;2^2  _  I2y%2. 

Arrange  the  expression  according  to  descending  powers  of  x ;  we 
then  have 

x^  +  2x^(2y^  -  3^2)  +  4^4  +  9^4  _  I2y^z\ 

that  is,  X*  +  2  a;2(2  «/2  -  3  z^)  +  (2  ?/2  _  3  ^2)2, 

which  is  {^2  +  (2  r  -  3  2:2)}2. 

Thus 

V(X*  +  4  2/4  +  9  5!4  +  4  a:2?/2  _  6  ^202  _  12  y2^2)=,(a;2  _|.  2  2/2  _  3  02). 

Again,  to  find  the  square  root  of 

a2  +  2  a6x  +  (62  +  2  ac)x2  +  2  bcx^  +  c2aj*. 
Arrange  the  expression  according  to  powers  of  a  ;  we  then  have 
a2  +  2  a(6a:  +  cx2)  +  62^2  +  2  bcx^  +  c^x:^, 
that  is,  a2  +  2  a(6x  +  cx2)  +  (6x  +  cx2)2, 

which  is      ^  {a  +  (hx  +  cx2)}2. 

Hence  the  required  square  root  is  (a  ■\-hx  +  cx^) . 
And  to  find  the  square  root  of 

x6_2x5  +  3x4  +  2  x\y  -  1)  +  x2(l  -  2  y)  +  2  xy  +  y^. 

The  expression  only  contains  y'^  and  y  ;  we  therefore  arrange  it 
according  to  powers  ofy;  we  then  have 

?/2  +  2  y(x^  -  x2  +  x)+  x6  -  2x5  +  3  x*  -  2x3  +  x2. 

Now  if  the  expression  is  a  complete  square  at  all,  the  last  of  the 
three  terms  must  be  the  square  of  half  the  coefficient  of  y ;  and  it 
is  easy  to  verify  that 

x6  -  2  x^  +  3  X*  -  2  x3  +  x2  is  (x«  -  x2  -f  x)2. 

Thus  the  given  expression  is 

?/2  +  2  y{x^  -  X2  +  X)  +  (X3  -  X2  +  X)2. 

The  required  square  root  is  therefore 
y  +  x^  —  x2  +  X, 


POWERS  AND  ROOTS.  313 

From  the  above  it  will  be  seen  that  however  many 
terms  there  may  be  in  an  expression  which  is  a  perfect 
square,  the  square  root  can  be  written  down  by  inspection, 
provided  only  that  the  expression  contains  only  two  difs 
ferent  powers  of  some  particular  letter. 

EXAMPLES  LXIII. 
Write  down  the  square  roots  of  the  following  expressions. 

1.  9 a;2  _  30 xy  +  25 2/2.  5.   x^  -6x^y^  +  9y\ 

2.  25  a;4- 30x22/2  + 9  y4.  6.   9x12-6x62/3  +  2/6. 

3.  4  X*  -  12  x22/2  +  9  2/*.  7.   ^  x^  -  ^  x^  +  i  J/«. 

4.  4x10-12x62/8  +  92/6  8.    1  x^  -  I  x^i/' +  i. 

9.  25  X82/6  _  40  a268x42/8  +  16  a'^b\ 

10.  ^+8x22/3  4.  i6a22/6. 
a2 

11.  9^-24x22/'«  +  16^^ 

a2  52 

12.  ^_42^+49aio. 
x6  x8 

13.  a2+462  +  9c2+ I26c  +  6ca  +  4a6. 

14.  4  a2  +  62  +  9  c2  +  6  6c  -  12  ca  -  4  ab. 
16.  4  a*  +  6*  +  c*  -  2  62c2  _  4  c2a2  +  4  a^b^. 

16.   25  a*  +  9  6*  +  4  c*  +  12  62c2  +  20  c2a2  -  30  a^b"^. 

212.  In  order  to  show  how  to  find  the  square  root  of 
any  algebraical  expression,  we  will  take  an  expression 
and  form  its  square,  and  then  show  how  to  reverse  the 
process. 


314  POWERS   AND   ROOTS. 

Consider,  for  example, 

x'  +  2xy  +  3f (i.), 

whose  square  is 

x^  +  4.:ty-\-10o^y^-{-12xf-[-^y^    .     .     .     (ii.), 

both  expressions  being  arranged  according-  to  descending 
powers  of  x. 

We  may  write  the  square  of  x^  -{■  2  xy  -\- ^  y^  in  either 
of  the  following  forms  : 

[o?  -^  {2xy  +  ^f)Y  =  x'  ^-2x\2xy  ^^y'') 

j^{2xy^Sfy (iii.), 

{{:>?  +  2xy)-\-  ^f\' ^{0^  +  2xyy  +  2{a? ^-2xy)^f 

•\-{^fY (iv.). 

Now  it  is  clear  that  the  first  term  of  (ii.)  is  the  square 
of  the  first  term  of  (i.).  Hence  the  first  term  of  the 
root  of  (ii.)  is  found  by  taking  the  square  root  of  its 
first  term. 

Again,  we  see  from  (iii.)  that  when  we  have  sub- 
tracted x^  (the  square  of  the  first  term  of  the  root),  the 
term  in  the  remainder  which  contains  the  highest  power 
of  a;  is  2  ic^  X  2  xy,  which  is  twice  the  product  of  the  first 
and  second  terms  of  the  root. 

Hence,  after  subtracting  from  (ii.)  the  square  of  the 
first  term  of  the  root,  the  second  term  is  obtained  by 
dividing  the  first  term  of  the  remainder  by  twice  the  first 
term  of  the  root. 

Again,  we  see  from  (iv.)  that  when  we  have  subtracted 
{x^  ■\-2xyy,  that  is,  the  square  of  the  part  of  the  root 
already  found,  the  term  in  the  remainder  which  contains 


POWERS   AND   ROOTS.  315 

the  highest  power  of  a;  is  2  cc^  X  3/,  which  is  twice  the 
product  of  the  first  and  third  terms  of  the  root. 

Hence,  after  subtracting  from  (ii.)  the  square  of  that 
part  of  the  root  already  found,  the  next  term  of  the  root 
is  obtained  by  dividing  the  first  term  of  the  remainder  by 
twice  the  first  term  of  the  root. 

If  we  now  subtract  the  square  of  x^  -\-2xy  -^-^y^  from 
the  given  expression,  there  will  be  no  remainder;  and 
hence  Qi?-{-2xy  -\-^y^  \^  the  required  root. 

We  will  now  consider  the  most  general  case. 

Suppose  we  have  to  find  the  square  root  of  {A  +  BY, 
where  A  stands  for  any  number  of  terms  of  the  root, 
and  B  for  the  rest )  the  terms  in  A  and  B  being  arranged 
according  to  descending  {or  ascending)  powers  of  some 
letter,  so  that  every  term  in  A  is  of  higher  {or  lower) 
dimensions  than  any  term  in  B. 

Also  suppose  that  the  terms  in  A  are  known,  and  that 
we  have  to  find  the  terms  in  B. 

Subtracting  A^  from  {A  -f  B)^,  we  have  the  remainder 
{2A  +  B)B. 

Now  from  the  mode  of  arrangement  it  follows  that  the 
term  of  the  highest  (or  lowest)  degree  in  the  remainder 
is  twice  the  product  of  the  first  term  in  A  and  the^rs^ 
term  in  B. 

Hence,  to  obtain  the  next  term  of  the  required  root, 
that  is,  to  obtain  the  highest  (or  lowest)  term  of  B,  we 
subtract  from  the  whole  expression  the  square  of  that  part 
of  the  root  which  is  already  fonnd,  and  divide  the  first 
term  of  the  remainder  by  twice  the  first  term  of  the  root. 

The  first  term  of  the  root  is  clearly  the  square  root  of 
the  first  term  of    the  given  expression;  and  when  we 


316  POWERS   AND   ROOTS. 

have  found  the  first  term  of  the  root  we  can  find  each  of 
the  other  terms  in  succession  by  the  above  process. 
For  ex'ample,  to  find  the  square  root  of 

xi  ^4x^y  -^  10  xV  +  12  xy^  +  9  y^. 

x*  +  4 x3y+ 10 x'^y^+12 xy^-^9y^  (pfi-\-2xy+  3 y^ 

(X2)2  =  ^^  *~ 


(x2  +  2  xyy  =  x*  +  4  x^y  +  4  x^y^ 
(x2  +  2xy  +  3?/2)2  =  x^-{-4.x^y+  lOxV  +  12x^3  ^ 


The  given  expression  must  first  be  arranged  according  to  ascend- 
ing or  descending  powers  of  some  letter. 

We  then  take  the  square  root  of  the  first  term  of  the  given 
expression  :  we  thus  obtain  x2,  the  first  term  of  the  required  root. 

Now  subtract  the  square  of  x2  from  the  given  expression,  and 
divide  the  first  term  of  the  remainder  by  2  x^ :  we  thus  obtain  2  x.v, 
the  second  term  of  the  root. 

Now  subtract  the  square  of  x2  +  2  xy,  which  is 
X*  +  4  x3?/  +  4  x2?/2, 

from  the  given  expression,  and  divide  the  first  term  of  the 
remainder,  namely  6  x^y"^^  by  2  x2 :  we  thus  obtain  3  !/2,  the  third 
term  of  the  root. 

Subtract  the  square  of  x2  +  2'xy  +  3?/2  from  the  given  expression, 
and  there  is  no  remainder. 

Hence  x^  -\-  2  xy  -\-  S  y^  is  the  required  square  root. 

The  squares  of  x2,  x2  -f  2  xy,  etc. ,  are  placed  under  the  given 
expression,  like  terms  being  placed  in  the  same  column:  the 
remainder  left  after  taking  away  any  square  is  then  obvious  by 
inspection. 

213.  Instead  of  finding  each  square  independently, 
some  labour  can  be  saved  by  making  use  of  the  previous 
square.  Thus  the  process  of  finding  the  square  root  of 
an  algebraical  expression  is  generally  written  as  follows, 
the  same  example  being  taken  as  before : 


POWERS   AND   ROOTS.  317 

a^  +  4  x^y  +  10x2y2  +  12  icy3  +  9  y4  (x2  +  2  xy  +  3  y2 


2x2  +  2ic?/)  4x^y 

4  ic^y  +    4  x2y2 

2x2  +  4xy +  3?/2)  6x22/2 

6x2?/2+  12x?/3  +  92/* 

The  first  term  of  the  root  is  x2.  Having  subtracted  the  square  of 
x2,  namely  x*,  the  first  term  of  the  remainder  is  4  x^y. 

Now  double  the  part  of  the  root  already  found  and  divide  the 
first  term  of  the  remainder  by  the  result ;  we  thus  obtain  the 
next  term  of  the  root,  namely  2xy.  Add  this  term  of  the  root  to 
2  x2,  placing  the  sum  in  the  ordinary  position  for  the  divisor.  Now 
multiply  the  sum  by  2xy  and  subtract  the  product  from  the 
remainder  4x^y -\- '■■,  we  then  have  the  remainder  Qx^y^+---. 
Repeat  this  process  as  often  as  may  be  necessary. 

[It  is  of  importance  to  notice  that  by  the  above  process  we 
have  at  the  end  of  the  second  stage  subtracted  altogether  - 

(x2  +  2xy)2, 

since  (x2  +  2  xy)^  =  (x2)2  +  (2  x2  +  2  xy)  x  2  xy. 

Similarly  at  the  end  of  every  stage  it  will  be  seen  that  what  is  sub- 
tracted on  the  whole  is  the  square  of  that  part  of  the  root  which 
has  been  found  up  to  that  stage.] 

EXAMPLES  LXIV. 
Find  the  square  roots  of  the  following  expressions : 

1.  x*  +  2x8  +  3x2  +  2x+  1. 

2.  4x4  -8x8  +  4x+  1. 

8.  9x*- 36x8 +  72x4- 36. 

4.  1-xy  -  Y-  x2y2  4.  2  x3y8  +  4  x*y*. 

6.  4x*  +  4x8-  ^x  +  tV 

6.  x4-2x3  +  |x2- 1x  +  tV- 


318  POWERS   AND   ROOTS. 

7.  x*  +  2 x"'?/  +  SxV  +  2 X2/3  +  ?/4. 

9.  l  +  4x+ 10x2+ 12x3  + 9ic*. 

10.  4x*-4x3  +  3x2-x  +  ^. 

11.  (1  +  2x2)2  _  4x(l -x)(l+2x). 

12.  x'''-4x5  +  6x*-8x'^  +  9x2-4x  +  4. 

13.  9x8- 12x5  +  22x4  +  x2+ 12X  +  4. 

14.  x6  -  22  x*  +  34  x3  +  121  x2  -  374x  +  289. 

15.  a2_aja;+ ix2  +  8a-4x+ 16. 

16.  x8  +  2x7+  x6-4x5-  12 X*  -  8x^  +  4x2  +  16x+  16. 

17.  16x2 -96x  + 216-^  +  ^- 

X  X2 

18.  x6-6x*  +  15x2-20  +  lf--^+l. 

*  X2         X*        X^ 

19.  X*  +  2  x3(y  +  0)  +  x2(?/2  +  2.2  +  4  2/0)  +  2  x?/0(?/  +  z)  +y^z^. 

20.  2x2(?/  +  0)2  +  2l/2(2;  +  x)2  +  2  02(ic  +  2/)2  +  4a;y2(X  +  2/+0). 


FBACTIONAL  AND  NEGATIVE  INDICES.         319 


CHAPTER  XXL 
Fractional  and  Negative  Indices, 

214.  We  have  hitherto  supposed  that  an  index  was 
always  a  positive  integer ;  and  this  is  necessarily  the  case 
so  long  as  we  retain  the  definition  of  Art.  19 ;  for,  with 
that  definition,  such  an  expression  as  a^,  for  example, 
has  no  meaning  whatever. 

It  is  however  found  convenient  to  extend  the  meaning 
of  a",  so  as  to  include  fractional  and  negative  values  of  n. 

Now  it  is  essential  that  algebraical  symbols  should 
always  obey  the  same  laws  whatever  their  values  may  be ; 
we  therefore  do  not  begin  by  assigning  any  particular 
meaning  to  a",  when  n  is  not  a  positive  integer,  but  we 
first  impose  the  restriction  that  the  meaning  of  a"  must 
in  all  cases  be  such  that  the  fundamental  Index  Law, 
namely, 

a"^  xa''^^  a'"+'', 

shall  always  he  true ;  and  it  will  be  found  that  the  above 
restriction  is  of  itself  sufficient  to  define  the  meaning  of  a" 
in  all  cases,  so  that  there  is  no  further  freedom  of  choice. 

For  example,  to  find  a  meaning  of  a^  consistent  with  the  Index 
Law. 

We  must  have 

a^  X  a^  =  a~    ^  =  a^  =  a. 

Thus  a*  must  be  such  that  its  square  is  a,  that  is  a*  must 
mean  y/a. 


320  FRACTIOISrAL   AND  NEGATIVE  INDICES. 

Again,  to  find   a  meaning  of  a~i  consistent  with  the  Index 
Law. 

We  must  have  a^  x  a~^  =  a^~^  =  a^ ; 

a 

Thus  a-i  must  mean  — 
a 

We  now  proceed  to  consider  the  most  general  cases. 


215.    To  find  the  meaning  of  a",  where  n  is  any  posi- 
tive integer. 

By  the  index  law 

111 
a^  X  a""  X  a""  X  •  •  •  to  n  factors 


1 
Hence  a"  must  be  such  that  its  nth  power  is  a,  that  is 

1 
a"  =  ^a, 

216.   To  find  the  meaning  of  a",  where  m  and  w  are 
any  positive  integers. 
By  the  index  law 

m  m  m 

a"  X  a**  X  a"  X  •  •  •  to  n  factors 

™  +  ^  +  ^  4- ...  to  n  terms  ^ 


Thus  a"  is  equal  to  the  nth  root  of  a"^,  that  is 


FRACTIONAL   AND   NEGATIVE   INDICES. 

We  have  also 

111 
a"  X  a**  X  a"  X  •  •  •  to  m  factors 


321 


=  a 


'-  +  '-  +  '-  + 

~      n     n 


a". 


Hence  a**  may  be  considered  as  the  mth  power  of  a",  and 
by  Art.  215, 

1 
a"  is  -l^a 

m 

Thus  we  may  consider  that  a**  is  the  nth  root  of  the  mth 
power  of  a,  or  that  it  is  the  mth  power  of  the  nth  root  of  a ; 
which  we  express  by 

a"  =  V  («")  =  (  V")" 

Note.  —  It  should  be  remarked  that  it  is  not  strictly  true  that 
^(a"*)={;y(a)}"'  unless  by  the  wth  root  of  a  quantity  is  meant 
only  the  arithmetical  root.  For  example  y/{a^)  has  two  values, 
namely  ±  a^,  whereas  (  ^ay  has  only  the  value  +  a^. 

Examples 

(i.)  8^  =  (^8)2  =  22  =  4,  (ii.)  4^  =  {/48  =  ^64  =  8, 

(iu.)  3^=^35=^43. 


217.   To  find  the  meaning  of  a^. 
By  the  index  law 


a''xa'^=  a°+'" 


a™; 


.-.  a**  =  a^a"*  =  1- 
Thus,  whatever  a  may  he,  a^  =  1. 


322  FRACTIONAL   AND   NEGATIVE   INDICES. 

218.    To  find  the  meaning  of  a~"*,  where  m  has  any 
positive  value. 
By  the  index  law 

and,  by  the  preceding  article,  a*'  =  1 ;  -a 

.'.  a"*  X  a-"*=  1, 

so  that  a-"*  =  — ,   and  a""  =  —^ 

Thus  we  can  transfer  any  factor  from  the  numerator  to 
the  denominator,  or  from  the  denominator  to  the  numerator, 
of  a  fraction,  provided  we  change  the  sign  of  its  index. 

For  example  2-2  =  —  =  1, 

-  =  a6  1  =  — -» 
and  ^  =  a^b'^x-^y-^  = 


219.  In  the  preceding  articles  we  have  found  that 
in  order  that  the  fundamental  Index  Law,  namely 
«»*  X  a"  =  a'"+**,  may  always  be  obeyed,  a"*  must  have  a 
definite  meaning  when  n  has  any  given  positive  or  nega- 
tive value.  It  can  be  proved  ^'^  that,  with  the  meanings 
thus  obtained, 

a'"  X  a~  =  a'"+",  (a"*)"  =  a"*",  and  (aby  =  a'*6*», 
are  true  for  all  values  of  m  and  n. 

It  therefore  follows  that  we  may  deal  with  quantities 
containing  fractional  or  negative  indices  in  precisely  the 
same  way  as  if  the  indices  were  positive  integers. 

*  See  Treatise  on  Algebra,  Art.  168. 


FRACTIONAL   AND   NEGATIVE  INDICES.         323 

For  example  {ah^-y  =  a^^'^ft^''^  =  ah. 


a^  xa^  =  a^  ^  =  a*  =  «/a. 

1  1  _2  '+i-2 

a^  X  a^  y-  a  ^  =  a-   ^  3  —  ^o  =  i. 

a* 
a-2/a-*  =  a-2-(-'*)  =  a^. 

^(a86-8c*)  =  ^(a3)  .  ^(6-3)  •  ^(c*)=  a^&"M  =ah-^c\ 

EXAMPLES  LXV. 
Find  the  numerical  values  of 

1.  si        3.   16-^.         6.  (^\)-i      7.  (27)-i         9.    i^:^^)-^. 

2.  rl      4.    (^V)'^-     6-  (fl)"^-      8-  (100)-^.     10.    (t?^)-! 
Simplify 

11.  a^  X  a"i  14.   a  x  a"i  17.    (a^ft"^)^. 

12.  a^  X  J.  15.   a^6"^  x  a~h\      18.    (a^6'^)«. 

13.  a~^  X  a.  16.    (a^6)2.  19.    (a~h^)^. 

20.    (a*6"^)l  ^''^-    (^'-0^  X  (^-^)' X  C^'"')*"- 

i  28.  ^-^^. 


21.  {(a-2)2}t. 

22.  {(a"bT^. 

23.  {(a"^)8ri  ^^ 


80. 


(xyzy+v+' 


24.  a^  X  a  -f  X  a^^.  x^+'j/^+^z^+y 

25.  a'  X  a"^  x  (a2)-H«T^)*.  31.    (a^^C^h^  ^  (aM&^. 

26.  a:P+«  x  x^-«  h-  x'^p.  {b^)^a^  (p^y 


324         FRACTEONAL   AND   NEGATIVE  INDICES. 


a2y3 


33.     X^P+<1  X  XP-"^  X  (a;2)9-2'-  ^  CC^-8r. 

r     P-?W     ff^ir-kpj-     r-p'\q  p-g      q-r      r-p 

Z^.   \xP^  )\x^'  f\x^  I  -^x  "-        P        «  . 

|-r    P2i:\?Z!!l'!:^  p-g      9-r     r-j> 

35.    Llic  '^    M  J  '    XX  '•       !>  "^  «  . 
Express  with  fractional  indices  and  simplify 
36.    >  X  >  X  >.  4o_    ^^,  ^  ^^,  ^  ^-i  ^  ^1 

-37.    ^(a^y)x^(a2/2).  ^^^    ^a^^{aV(«-^^)}- 

38.  ^(a2^5)xv(a3x).       ^  ^^^    ^(a26i0c5)^^a66,3). 

39.  ^a^  x  ^a^  x  a"^  --  a^i".  43.    V(«^^"^)^  ^(a-*&^). 

44.    ^(a^feM)  x  6"^  X  (cV)"^. 
Express  the  following  quantities  with  radical  signs  and  positive 
indices. 

45.  J  -  a-\  47.   a^6-i  -  ah'^. 

46.  a-46"^.  48.    a-3&"^  +  3-2a3V^. 


220.  In  the  following  examples  the  foregoing  princi- 
ples are  applied,  but  the  results  of  the  operations  of 
addition,  etc.,  which  have  to  be  performed  on  the  indices 
are  given  without  any  of  the  intermediate  steps. 

Ex.  1.   Multiply    a^  +  1    +  a~^  by  a^  -  1  -I-  a^ 
a^  +  1    +a'^ 
a^  -  1    +  a~^ 


a^  -^a^ 


1  _i 

a^  -\-  a  3 


+  1  +  a"^. 


FRACTIONAL  AND   NEGATIVE   INDICES. 


325 


Ex.  2.   Divide  a+b  +  c  -Sa^b^c^  by  a^  +  b^  +  c^. 
We  perform  the  work  by  the  synthetic  method  [Art.  83.]. 


f 
(6^  +  c^) 


0  -Sahh^  +(6  +  c) 

a3  (6^  +  c3)  +  a^(b^  +  c^  +  2  b^c^)  -(b  +  c) 


aJ(^b^  +  c^)  +  6^  +  c3  -  dM 


0. 


Ex.  3.     Find  the  square  root  of 


•t  4-  rf. 


ic^  -  4  X »  +  2  x^  +  4  a;  -  4  X  5  +  jc 
4 


^  -  4  x^  +  2  x^  +  4  X  -  4  x^  +  x^  (x^  -  2  x^  +  a;^ 


(X^)2  =  X^ 

(x^  -  2x^)2  =  x^  -  4x^  +  4x 
(x^  -2  x^+x^)2  =  x^  -  4  x^  +  2  x^  +  4  X  -  4  x^  +  x^. 

EXAMPLES  LXVI. 


Multiply 


1.  x^  +  y^  by  x^  -  y^. 


^_.A 


4.   x^  +  y3  by  x^  -  yi 


S.  x^  +  y^  by  x^  -  y^. 


6.   x^  +  x"^  +  1  by  x^  -  1. 
3.   xf  +  1  by  x^  -  1.  6.   x^  -  x^^y^  +  y^  by  x^  +  y^. 

7.  a^  +  a^ft^  +  b^  by  a^  -  b\ 

8.  a^  +  a^&^  +  6^  by  a^  -  ah^  +  h^ 

5  3  1  _1  3  1 

9.  x^  —  X*  +  x^  —  X  ^  by  x^  4-  x^. 

10.  x^  -  x^  +  x^  -  X  by  x^  +  ic^. 

n  _n  , 

11.  x«  +  x2  + 1  by  x-«  +  x  2  4-1. 

12.  x^  +  x"y"  +  y2n  by  a;-2n  _^  x-"y~"  4-  y~^. 

13.  a^  +  6^  4-  c^  -  bh^  -  c^a^  -  ah^  by  a^  4-  6^  4-  ci 


326  FRACTIONAL   AND   NEGATIVE  INDICES. 

14.  ^J-  tV  a&^  +  tV  «^&  -  2V  ^^  by  I  «"^  +  i  ^^' 

15.  81  x^  -  27  x^y^  +  9  x^y 3  _Sx^y  +  y^  by  3  x^  +  yi 
Divide : 

16.  a-b  by  a^  -  bK  19.  x^  +  y^  by  a;^  +  yi 

17.  x^-y^hyx^-yK  20.   a;  -  243  ?/ 3  by  x^  -  3  2^^. 

3n  3n  n  n 

18.  x'^  -  y^  hj  x^  -  y^,  21.    ?/*4-&V+&*  ty  y-6^y^+6. 

22.  x^y~^  +  2  +  x"%^  by  x'fy"^  -  1  +  x~^y^. 

23.  a^  +  a^b^  -  ah^  -  ab  +  ah^  +  b^  by  a^  +  b\ 

1  _4  2  _2 

24.  a3  -  2  4-  a  3  by  a3  -  a  3. 

25.  (a;  -  x-i)  -  2(x^  -  x"^)  +  2(x^  -  x"^)  by  x^  -  x^. 

26.  x^^/""^^  +  y~^'x~^  by  x^y"^  +  y^x~^. 

27.  Simplify  (aft-^cS)^  x  (a^&^c-i)^  x  (a-^bc^)^. 

28.  Simplify  (a  -  4  b^  (a  +  2  a^&^  +  4  63)  (a  -  2  a^6^  +  4  5^). 

29.  Show  that  1  -  a^  +  b^-a-^-b-^  ^  (g  _  a-i)(6  _  5-1). 

2 

30.  Show  that  -^ —- p—  +  -^ —  =  2  +  xl 

x^  -  1     x^  +  1     x^  -  1     x^  +  1 

31.  Multiply  4x2 -5x-4- 7x-i  +  6x-2  by  3x-4  +  2x-i, 
and  divide  the  product  by  3  x  —  10  +  10  x~i  —  4  x~^. 

32.  Simplify  — ^—  +  ^ —  +  — ^  +     ^ 


1-x^     l  +  x^     1+x^     1+^ 
33.    Multiply 
(a  +  6) ^  +  (a2  _  62)i  +  (oj  _  5)f  by  (^  +  6)i  _  («  _  6)i 


FRACTIONAL  AND   NEGATIVE   INDICES.         327 

34.  Write  down  the  square  roots  of 

(i.)  a;^  +  2  x^  +  1,  (ii.)  4x^  -  4:X^y^  +  y^, 

(iii.)  ax^  -  2  aV  +  a^x. 

35.  Find  the  square  root  of 

4a;2a-2  -  12xa-i  +  25  -  24a;-ia  +  16x-^aK 

36.  rind  the  square  root  of 

25  a;2y-2  +  ^  y^x-"^  -  20  xy-i  -  2  yx-^  +  9. 

37.  Find  the  square  root  of 

x^  —  2  a~^x'^'  +  2  a'^a;'  +  oT^x^  —  2  a^x^  +  a^. 

88.  Find  the  square  root  of 

x^  -  4x^^  4-  4a;  +  2a;^  -  4a;*  +  xi 
Solve  the  following  equations : 

89.  a;^-2x^  +  l=0.  41.   a;^  -  26  a;^  -  27  =  0. 
40.   a;  -  8a;^  +  12  =  0.                   42.   x^  +  Sx~^  =  4. 

48.  4a;^-3a;"^  =  4. 


328  SURDS.      COMPLEX  QUANTITIES. 


CHAPTER  XXII. 

Surds.    Complex  Quantities. 

221.  Definitions.  A  surd  is  a  root  of  an  arithmetical 
number  which  can  only  be  fouDd  approximately. 

Thus  ^2  and  ^4  are  surds. 

An  algebraical  expression  such  as  -^/a  is  also  often 
called  a  surd,  although  a  may  have  such  a^value  that  ^a 
is  not  in  reality  a  surd. 

Surds  are  said  to  be  of  the  same  order  when  the  same 

root  is  required  to  be  taken.     Thus  ^2  and  6^  are  called 
surds  of  the  second  order,  or  quadratic  surds ;  also  ^4 

and   5     are   surds  of  the   third   order,   or  cubic  surds; 
and  -^5  is  a  surd  of  the  nth.  order. 

Two  surds  are  said  to  be  similar  when  they  can  be 
reduced  so  as  to  have  the  same  irrational  factors. 

Thus  2,y/2  and  6^2  are  similar  surds. 

The  rules  for  operations  with  surds  follow  at  once 
from  the  principles  already  established. 

Note.  —  It  should  be  remarked  that  when  a  root  symbol  is 
placed  before  an  arithmetical  number  it  denotes  only  the  arith- 
metical root,  but  when  a  root  symbol  is  placed  before  an  algebraical 
expression  it  denotes  any  one  of  the  roots.  Thus  -^a  has  two 
values,  but  y/2  is  only  supposed  to  denote  the  arithmetical  root, 
unless  it  is  actually  written  ±  y/2. 


SURDS.      COMPLEX  QUANTITIES.  329 

222.   Any  rational  quantity  can  be  written  in  the  form 
of  a  surd. 

For  example 

and  2  =  ^2  _  ^23  =  «/2«. 

Again  -^2  =  ^l^2'\  =  ^2'. 

Also,  since  -^a  x  ^b  =  -Vab,  we  have 

3V2  =  V3'  X  V2  =  V(3'  X  2)  =  V18, 
2^5  =  ^2^  X  -^5  =  ^(2^  X  5)  =  ^40, 
and          a^yft  =  ^a«  X  V^  =  ^^^• 
Conversely,  we  have 

V18  ^  V(9  X  2)  =  V^  X  V2  =  3  V2, 

^135  +  ^40  =  ^(3^  X  5)  +  ^(2«  X  5) 
=  3^5  +  2^5  =  5^5, 
and  -^a^  4-  Va6^  =  ay'ct  4-  6^a  =  (a  +  6)  y'a. 


t.   -4ny  ^i«o  surds  can  be  reduced  to  surds  of  the  same 
order. 

This  follows  at  once  from  the  fact  that  -y/a  =  "^'a*". 
For  example,  to  reduce  ^2  and  ^5  to  surds  of  the 
same  order. 
We  have 

-2/2  =  ^1^2^  =  ^2«,  and  ^5  =  ^l^5'\  =  ^5^; 

thus   the  equivalent  surds  of   the  same  order  are   ^8 
and  ^25. 

Again,  to  reduce  ->/a  and  ^b  to  equivalent  surds  of 
the  same  order. 


330  SURDS.      COMPLEX  QUANTITIES. 

We  have 

^a  =  ^{^a^^l  ='Va^  and  ^b  =  Vf  V^"i  ="76". 
Hence  the  required  surds  are  "4ya'"  and  ''^b''. 

Ex.  1.    Reduce  ^3  and  ^5  to  surds  of  the  same  order. 
The  L.  C.  M.  of  4  and  6  is  12.     Hence  we  have 

^3  =  ^{^33}  =1^27,  and  ^5  =  ^{^62}  =1^25. 
Thus  ^^27  and  1^5  are  the  required  surds. 

Ex.  2.  Which  is  the  greater,  ^14  or  ^6  ?  We  must  reduce  the 
surds  to  equivalent  surds  of  the  same  order. 

Thus  '      ^14=^142=^196, 

and  ^6  =  ^63  =  ^216. 

It  is  now  obvious  that  y/6  is  greater  than  ^14. 

224.  The  product  of  two  surds  of  the  same  order  can 
be  written  down  at  once  from  the  formula 

^a  x^b  =  Vab. 

When  surds  are  of  different  orders  their  product 
cannot  be  simplified  until  they  are  reduced  to  equivalent 
surds  of  the  same  order. 

Ex.  1.   Multiply  V5  by  V20. 

We  have      ^6  x  V20  =  VC^  X  20)  =  VlOO  =  10. 

Ex.  2.   Multiply  y/2  by  ^3. 

V2  X  ^3  =  ^3  X  ^32  =  ^(28  X  32)  =  4/72. 

Ex.  3.   Divide  ^2  by  V^- 

^2/  V6  -  ^27  ^6-3  =  ^(22/63)  =  ^3^j. 


SUKUS.      COMPLEX   QUANTITIES.  331 

Ex.  4.  Multiply  4  ^3  +  4^2  by  2  V3  -  2  ^2. 

The  process  is  as  under : 

4V3  +  4V2 
2V3-2V2 

8x3  +  8^/6 

-  8y/6  -  8  X  2 

24  -16      =8. 

EXAMPLES  LXVII. 
Simplify 

1.  V27  +  >/48.  15.   ^i2xV24. 

2.  V50  +  V98.  16.  2Vfx3v^ 

3.  V45  +  2V126.  ^^  3^ 

4.  2V180-V406.  ^         ^ 

6.    2V28-V63.  "•    V12XV27XV75. 

6.  5V2-08-3V3-26.  "'  ^^«  ><  ^«  ><  ^• 

7.  3  3.5  +  ^625.  ^-  ^12x^75x^30. 

8.  3^72  -  2  </243.  ^l.  </6  x  ^12  x  ^18. 

9.  4^448-15^7.  22.  VlO  x  ^00. 

10.  V512  -  V^O  -  y/9S.  23.    ^4  X  ^. 

11.  3  V12  -  V27  +  2  V75.  24.    y/m-^y/50. 

12.  V147-2V27-V3.  26.    V63^Vn2. 

13.  5  Vi  -  \/8  +  -^-  26.    V20  X  y/9Q  -  ^30. 

■v/18 

14.  y/15  X  V60.  27.    ^147  -f-  ^35  x  ^736. 

28.  Which  is  the  greater,  ^3  or  ^(5^)? 

29.  Arrange  in  order  of  magnitude  ^^50,  ^^344,  and  ^402. 

30.  Multiply  V6  -  V3  hy  y/Q  +  ^3. 

31.  Multiply  2  V5  +  3  ^3  by  3  V5  -  4  V3- 

32.  Multiply  ^2  +  ^3  +  V^  by  2  v2  +  3  V3  +  v'B. 

33.  Multiply  1  +  ^3  +  ^6  by  1  +  ^3  -  ^5. 


332  SUKDS.      COMPLEX  QUANTITIES. 

34.  Find  the  square  of  ^2  +  V^  +  V^- 

35.  Find  the   continued  product  of  ^2  +  y'3  +  ^5,   —^-\-^Z 

+  V5,  V2  -  V3  +  V5,  and  V2  +  V^  -  V^- 

225.  When  surds  occur  in  the  denominators  of  frac- 
tions they  can  be  got  rid  of,  and  the  denominators  are 
then  said  to  have  been  rationalized. 

The  following  examples  will  sufficiently  illustrate  the 
process : 

3    ^    3xV5    ^3    .^ 

3V2^3V2x  V7^3    ,^^ 

2+ V5^(24-V5)(V5  +  l)^2V5  +  5  +  2+V5 
V5-1      (V5-1)(V54-1)  5-1 

=  i(7  +  3V5), 

a-^b^  (a--y/b)  (a  -  ^b)  ^a^-2a^b-\-b 

It  is  important  to  notice  that  a  ±  ^b  is  made  rational 
by  multiplying  by  a  if  ^b ;  also  that  ^a  ±  ^b  is  made 
rational  by  multiplying  by  -y/a  T  ^h. 

When  the  denominator  of  a  fraction  is  rationalized  its 
numerical  value  can  be  more  easily  found. 

226.  The  following  is  an  important  proposition. 

If  a-\-  ^b  =  «  +  ^^,  where  a  and  a  are  rational,  and 
^b  and  -^(S  are  irrational;  then  will  a=  a,  and  b  =  p. 

For  we  have       (a  —  a)+  -^b  =  ^fi. 


SUEDS.      COMPLEX  QUANTITIES.  333 

Square  both,  sides ;  then,  after  transformation,  we  have 
2(a  -a)-y/b  =  l3-  b-  {a  -  a)'. 
Hence,  unless  the  coefficient  of  ^h  is  zero,  we  must  have 
an  irrational  and  a  rational  quantity  equal  to  one  another, 
and  this  is  impossible. 

The  coefficient  of  ^h  in  the  last  equation  must  there- 
fore be  zero ;  hence  a  =  a.  And  when  a  =  a  we  have 
from  the  given  relation,  ^h  =  ^p,  ox  b  =  ft. 

Hence,  if  the  sum  {or  difference)  of  a  rational  quantity 
and  a  quadratic  surd  be  equal  to  the  sum  {or  difference)  of 
another  rational  quantity  and  a  quadratic  surd,  the  two 
rational  quantities  must  be  equal  to  one  another,  as  also 
the  two  irrational  quantitie,s. 

Note.  —  It  should  be  noticed  that  when  a  -f  ^/b  =  a  +  ^^/S  we  can 
only  conclude  that  a  =  a  and  6  =  /3  provided  that  y/b  and  ^/^  are 
really  irrational.  We  cannot,  for  example,  from  the  relation 
S  +  y/i  =  2  +  y/9,  conclude  that  3  =  2  and  4  =  9. 

227.  The  square  root  of  a  binomial  expression  which 
is  the  sum  of  a  rational  quantity  and  a  quadratic  surd 
can  sometimes  be  found  in  a  simple  form.  The-  process 
is  as  uncjer. 

To  find  V(^  +  V^)'  where  ^b  is  a  surd. 

Let  ^{a-\-^b)=^a-\--^/ft (i.) 

Square  both  sides ;  then 

Now,  since  -^/b  is  a  surd,  we  can  equate  the  rational  and 
irrational  parts  [Art.  226]  ;  hence 

«=«+^l (ii.) 

6  =  4a/J    J 


334  SURDS.      COMPLEX  QUANTITIES. 

Hence  [Art.  183]  a  and  ^  are  roots  of  the  equation        * 

a;2  _  aaj  +  -  =  0 ; 
and  these  roots  are  ; 

Thus  v(^+v^)=V{-'"^f~-j;.,,^^^    ^' 

It  is  clear,  that  unless  -yjio?  —  h)  is  rational,  the  right 
side  of  (iii.)  is  much  more  complicated  than  the  left. 
Thus  the  above  process  is  of  no  utility  unless  a^  —  &  is  a 
square  number ;  and  as  this  condition  will  not  often  be 
satisfied,  the  process  has  no  great  practical  utility. 

From  (ii.)  we  see  that  we  have  to  find  two  numbers 

whose  sum  is  a  and  whose  product  is  - ;  and  if  two  rational 

numbers  satisfy  these  conditions,  they  can  generally  be 
found  at  once  by  inspection. 

Ex.  1.   Find  y/{\b  +  2  V^B). 

Let  V(15  +  2  V56)  =  V«  +  V^- 

Square  both  sides  ;  then 

15  +  2  V56  =  a  +  ;8  +  2  ^afi. 
Equating  the  rational  and  irrational  terms,  we  have 
o  +  )8  =  15, 
o3  =  56. 
The  numbers  which  satisfy  these  relations  are  obviously  7  and  8. 
Hence  yj(l^  +  2  y/bQ)  =  V7  +  V^- 


-Tf-^        SURDS.   COMPLEX  QUANTITIES. 

^^t  --^x.  2.  Find  ^(6-^/35). 

\  I     Let  V(6  -  V35)  =  y/a-  y/$. 

...Square  both  sides  ;  then 

6  -  V35  =  a  +  iS  -  2  y/a$. 
:ce,  equating  the  rational  and  irrational  terms,  we  have 
a+)8  =  6, 


335 


Jy  mspection,  or  by  solving  the  equations  for  a  and  jS,  we  find  that 


EXAMPLES  LXVIII. 
Rationalize  the  denominators  of 


1 

3 

V7 

8. 

3^2 

2v3 

5          2 

1+V2 

2. 

2 

V5 

4. 

V6. 
V5 

6.    V2-1 

v^  +  1 

7 

2V5 
V6  +  V3 

11               1 

1+V2+V3 

8. 

15  +  14^3 
15-2V3 

12.             3 

2  +  V3  +  V5 

9. 

V6  +  3  ^3 
2  V6  -  V3 

13          ^        1        ^ 

V»/2  -  1  '    ^  +  1 

10. 

V6  -  3v/12 
2V6+  y/l2 

14.  ^        1        ^ 
^  -  1  \S/9  +  1 

15. 

Simplify 

1 

^         1 

(2-V3)2  '  (2+V3)2 

16. 

Simplify 

1 

/■ct 

,          1        . 

/0\«     '      /-O      1          *o\« 

(2-V3)«      (2+V3)» 

17.    Simplify  (3  +  V2)(5-v^) 
(3-v2)(6  +  V2) 


336  SURDS.      COMPLEX  QUANTITIES. 

18.  Simplify  2V15-3V5  +  2V2^ 

V15+V2 

19.  Simplify  (7-2V5)(5  +  V7)(31  +  13V5)^ 

(6-2  V7)(3+V5)(11  +  4V7) 

21.    Show  that  ,^1^0"^^^  =  2  +  V2  +  V3  +  V6. 


22.    Simplify 

1  ,  1 


+  ^ :: -^ 


1+V2  +  V3      -1+V2+V3      1-V2+V3      1+V2-V3 
24.    Show  that 

26.  Find  the  value  of  '^ — : —  to  three  places  of  decimals. 

V2-1 

27.  Find  the  value  of      "^       +  - — ^  to  three  places  of  decimals. 

2,    —   -y/O  2    +    -yjL 

Find  the  square  roots  of 

28.  6  +  V20.  32.   101  -  28  ^13. 

29.  16  +  6V7.  33.    117  +  36  VlO. 

30.  12-6V3.  34.   280  +  56V21. 

31.  28-5V12.  35.   4^  + 2  V2. 
Simplify 

36.  3V6-V2+V(7+2V10).        37.   6-4^3  +  ^(16-8^3). 


SUKDS.      COMPLEX  QUANTITIES.  337 

Find  the  square  roots  of 

38.  11  +  2(1+V5)(1+V7)-  40.    2a  +  2v(a2-cc2). 

39.  2x  +  2V(x^-l)-  41-   3a;-l+2V(2x2+x-6). 
Simplify 

42.    ^  ^  ^     ,  •  43 


V(16  +  6V7)  V(15  +  2V56) 

44.    V(7  +  2V10)  +  V(7-2V10). 
45     VC3  +  2V2)-V2. 

V2+V(3-2V2) 


2  + a/3 2-V3 

V2  +  V(2  +  V3)      V2  -  V(2  -  V3)' 


46. 

47.    Show  that V^  +  V^^ =  3  •  632 . .. 


48.  Find  the  value  of  x^  —  4  x  4-  5  when  x  =  2  +  y/b. 

49.  Show  that,  if  x2  =  a;  +  1,  then  x^  =  2x  +  1  and  x^  =  5x  +  3. 

50.  Show  that,  if 

x2  =  3  X  +  6,  then  x«  =  14  x  +  16  and  x*  =  67  x  +  70. 

COMPLEX  QUANTITIES. 

228.  In  the  chapters  on  quadratic  equations,  and  on 
equations  of  higher  degrees,  we  encountered  expressions 
which  involve  the  square  roots  of  negative  quantities. 
Such  square  roots  present  themselves  in  the  form  ^—c, 
in  which  c  is  a  positive  number.  This  is  called  a  pure 
imaginary  as  distinguished  from  the  mixed  binomial  form 
a-{-y/—c,  which  is  called  a  complex  quantity. 

Such  expressions  do  not  enter  largely  into  the  applica- 
tions of  elementary  algebraic  principles,  and  a  detailed 
study  of  them  would  displace  other  matters  of  more 
immediate  importance  to  the  student.     But  some  account 


338  SURDS.      COMPLEX   QUANTITIES. 

of  their  behaviour  in  the  operations  of  algebra  should  be 
given.  The  following  discussion  consists,  in  the  main, 
of  explanations,  without  formal  demonstration,  of  the 
principal  laws  of  operation  with  complex  quantities. 

229.  In  admitting  imaginary  forms  to  the  category  of 
algebraic  quantity,  it  becomes  necessary  to  enquire 
whether  the  algebraic  processes  may  be  applied  to  them 
without  limitation.  And  we  find  in  fact  at  the  outset^ 
that  if  we  attempt  to  apply  the  i-ule  that  the  sign  ^  is 
distributive  over  the  factors  in  a  product,  and  assert  that 


V— cxV  —  c  =  V(— c)  x{—c)  —  c, 
this  is  at  once  in  conflict  with  the  other  rule  that 

which  asserts  that  to  square   an  indicated  square  root 
has  merely  the  effect  of  removing  the  radical  sign. 

Driven  to  a  choice  of  interpretations,  we   reject   the 
former,  adopt  the  latter,  and  define  outright  that 

{y/—cy  =  -e, 

and  accept  the  other  consequent  interpretations  to  which 
this  definition  leads. 

230.   If  now  we  assume  the  commutative  and  associa- 
tive laws  in  multiplication  [Art.  52],  and  write 

ax-\/  —c  =  ^  —c  xa, 

(a  X  V—  c)  X  6  =  a  X  (V—  c  x  6), 
we  have 

Vc  X  V—  1  X  Vc  X  V—  1  =  Vc  X  \/c  X  V—  1 X  V^^, 


SURDS.      COMPLEX  QUANTITIES.  339 

whence         ( Vc  X  V—  1)^  =  —  c ; 
and  comparing  this  with 

we  see  that  it  is  legitimate  to  write 

V— c  =  VcxV^. 

The  form  Vc  x  V—  1,  which  is  the  one  universally 
employed,  is  usually  written  in  the  abbreviated  form  hi, 
in  which  i  now  takes  the  place  of  ^—1,  and  b  of  yc. 
The  symbol  ■\J—1,  or  its  equivalent  i,  is  called  the  imag- 
inary unit. 

231.  With  the  interpretation  oi  ■^J—l  we  have  now 
adopted,  we  have 

^«  =  _1,     V  =  -i,    ?•«=!,      etc., 
so  that  i,  —  1,  —  i,  1,  are  the  only  values  that  the  inte- 
gral powers  of  V~^  ^^^  assume. 

232.  For  the  imaginary  unit  in  its  combinations  with 
real  quantities  we  have  assumed  the  commutative  and 
associative  laws  in  multiplication,  namely 

(axi)xb  =  ax{i  X  b), 
and  it  is  equally  essential  that  the  like  laws  in  addition 
a  +  i=i-\-ay 

(a  +  i)  +  b  =  a-h{i-\-b), 
and  the  law  of  distribution 

i  x(a-{-b)  =  i  xa-{-i  xb 


340  SURDS.      COMPLEX  QUANTITIES, 

shall  also  obtain.  With  these  laws  and  the  interpreta- 
tion of  ^—  1  definitely  agreed  to,  we  can  perform  the 
fundamental  algebraic  operations  upon  complex  quanti- 
ties. [See  Treatise  on  Algebra^  pp.  221,  222 ;  also 
Chrystal's  Algebra,  Vol.  I.,  Chapter  XI.] 

The  following  examples  illustrate  the  application  of 
the  rules  of  addition,  subtraction,  multiplication,  and 
division. 

Ex.  1.   The  sum  of  a  +  hi  and  c  +  di  is 

{a-^U)^(c-\-di)  =  ia  +  c)  +  (h-\-d)i. 
Ex.  2.    The  difference  of  a  +  hi  and  c -{■  di  is 

{a  +  hi)-{c  + di)  =  {a-c)-^{h-d)i. 
Ex.  3.    The  product  of  «  +  hi  into  c  +  di  is 

(a  4-  hi)  (c  +  di)  =ac -{-  adi  +  hci  +  hdi^ 
—  {ac  —  hd)  +  {ad  -f  hc)i. 

Ex.  4.   The  quotient  of  a  +  hi  by  c  +  di  is 

a  +  hi  _{a  -\-  hi)  (c  —  di) 
c  +  di      (c  +  di)  (c  —  di) 

_ac  —  adi  +  hci  —  bdi^ 
c2  —  cdi  +  dci  —  d'H'^ 
_  (ac  +  hd)  +  (6c  —  acg)i 

C2  +  d2 

Ex.  5.   The  square  of  a  +  hi  is 

(a  +  hi)-^  =  a'^-b^-\-2  abi. 
Ex.  6.   The  reciprocal  of  a  +  hi  is 

1      _a  —  hi 
a  +  hi~  a^  +  b"^ 

From  these  examples  it  is  at  once  evident  that  sums, 
differences,  products,  and  quotients  of  complex  quantities 
are  themselves  complex  quantities.     It  may  likewise  be 


SURDS.      COMPLEX   QUANTITIES.  341 

shown,  that  from  the  other  algebraic  operations,  when 
applied  to  complex  quantities,  only  other  complex  quan- 
tities (or  possibly  reals  and  imaginaries*)  can  result. 
[See  Stringham's  Uniplanar  Algebra,  Chapter  III.] 

233.  When  two  complex  quantities  have  their  real 
parts  the  same  and  their  imaginary  parts  different  only 
in  sign,  they  are  said  to  be  conjugate  to  one  another.  For 
example,  —a  +  hi  is  conjugate  to  —  a  —  hi. 

Both  the  sum  and  the  ^woduct  of  two  conjugate  complex 
quantities  are  real. 

For,  let  the  expressions  be  a  +  hi  and  a  —  hi,  both  a 
and  h  being  real.     Their  sum  is 

a-\-hi-\-  a  —  hi  =  2  a, 
and  their  product  is 

(a  +  hi)  (a  -  hi)  =  a^  -  ahi  +  hai  -  hH^  =  a^  -f-  h^. 

Conversely,  if  the  sum  and  the  product  of  two  complex 
quantities  he  hoth  real,  the  complex  quantities  must  he  con- 
jugate. 

For,  let  the  expressions  be  a  +  hi  and  c  -f-  di,  a,  h,  c, 
and  d  being  real.     Their  sum  is 

a -\- hi -\- c -\- di  =  a -\- c -{- {h  ■\- d)  i, 
which  cannot  be  real  unless  6  -f  d  =  0 ;  and  their  prod- 
uct  is 

(a  +  hi)  (c  +  di)  =  ac  +  adi  -f-  hci  +  hdi^ 
=  {ac  —  hd)  +  (ad  -f  hc)i, 
which  cannot  be  real  unless  ad-{-hc  =  0. 

*  Reals  and  imaginaries  may  be  regarded  as  particular  cases  of 
complex  quantities,  the  former  being  complex  quantities  with  zero 
imaginary  parts,  the  latter  complex  quantities  with  zero  real  parts. 


342  SURDS.      COMPLEX  QUANTITIES. 

But  if  b  -\-  d—  0  and  ad-\-bc  =  0,  then  b{c  —  a)  =0, 
whence,  either  c  =  a,  or  6  =  0.     Then  : 

(1)  If  b  =  0,  from  b  -\-d  =  0  it  follows  that  d  is  also 
zero  and  the  original  expressions  are  real. 

(2)  If  c  =  a  {b  not  zero),  it  follows,  because  d  =  —  b, 
that  c  -\-di=  a  —  bi, 

which  is  conjugate  to  a  +  bi.  q.e.d. 

234.  When  an  expression  is  written  in  the  form  a  -j-  bi, 
it  is  understood  that  a  and  b  are  both  real. 

If  a-{-bi  =  0,  then  a  =  0.  and  6  =  0. 

For^  if  a  +  bi  =  0,  then  a-=  —  bi. 

But  a  real  quantity  cannot  be  equal  to  an  imaginary 
one  unless  both  are  zero. 

.-.  a  =  6  =  0. 

This  proposition  is  a  lemma  to  the  following. 

Two  complex  quantities  cannot  be  equal  to  one  another 
unless  their  real  and  their  imaginary  parts  are  respectively 
equal 

For,  if  a  4-  6i  =  c  +  di, 

then  a  —  c  =  {d  —  b)i'y 

whence,  as  previously  proved, 

a  — c  =  0,  d  —  b  =  0, 
that  is  a  =  c  and  d=b.  q.e.d. 


EXAMPLES  LXIX. 

Simplify  each  of  the  following  expressions  by  reducing  it  either 
to  a  form  having  no  imaginary  part,  or  to  a  form  having  no  real 
part ;  or  if  it  have  both  real  and  imaginary  parts,  reduce  it  to  the 
typical  form  a  +  bi. 


SUBDS.      COMPLEX  QUANTITIES.  343 


1-  h  *■  r^*-  ■>■ 

3i  1  -I 


m' 


^  5.    %±^-  8.    (l±i\*. 


1+i  2-5i  V  V'2  / 

3.  i±i.  6.  ii:^tO-'.  9  /^-^y. 

l-i  2i-l  '    \     I     / 

10.  (-3  +  20(3  +  20.  ^^      1  +  t  l-t 

11.  (5i  +  2)(15  +  60.  1-2?  l  +  2i 

12.  a  +  JV3)^.  15    2-3i  ^  2  +  3i 

13.  (i-^V3)«.  '   ^  +  ^*"  ^-2*'' 

16.  Write  the  equation  whose  roots  are  1,  i,  —  i,  and  —  1. 

17.  Write  the  equation  whose  roots  are    1,   \  +  \iy/Z,   and 

18.  li  ta=-\-\-\  1  ^3,  prove  that  1  +  w  +  «2  =  0. 


344  RATIO.      PROPORTION.      VARIATION. 


CHAPTER   XXIII. 
Ratio.    Proportion.    Variation. 

235.  Definitions.  The  relative  magnitude  of  two  quan- 
tities, measured  by  the  number  of  times  which  the  one 
contains  the  other,  is  called  their  ratio. 

Concrete  quantities  of  different  kinds  can  have  no 
ratio  to  one  another :  we  cannot,  for  example,  compare 
with  respect  to  magnitude  miles  and  tons,  or  shillings 
and  weeks. 

The  ratio  of  a  to  5  is  expressed  by  the  notation  a :  6  ; 
and  a  is  called  the  first  term,  and  b  the  second  term  of  the 
ratio. 

Sometimes  the  first  and  second  terms  of  a  ratio  are 
called  respectively  the  antecedent  and  the  consequent. 

It  is  clear  that  a  ratio  is  greater  than,  equal  to,  oi- 
less  than  unity,  according  as  its  first  term  is  greater  than, 
equal  to,  or  less  than  the  second. 

A  ratio  which  is  greater  than  unity  is  sometimes  called 
a  ratio  of  greater  inequality,  and  a  ratio  which  is  less  than 
unity  is  similarly  called  a  ratio  of  less  inequahty. 

236.  Magnitudes  must  always  be  expressed  by  means 
of  numbers ;  and  the  number  of  times  which  one  number 
is  contained  in  another  is  found  by  dividing  the  one  by 
the  other.     Hence 

a :  6  is  equal  to  — 
b 

Thus  ratios  can  be  expressed  as  fractions. 


RATIO.      PROPORTION.      VARIATION.  345 

237.  A  fraction  is  unaltered  in  value  by  multiplying 
its  numerator  and  denominator  by  the  same  number. 
[Art.  158.] 

Hence  also  a  ratio  is  unaltered  in  value  by  multiplying 
each  of  its  terms  by  the  same  number. 

Thus  the  ratios 

2:3,  6:9  and  2  m  :  3  m 
are  all  equal  to  one  another. 

Again,  the  ratios  4:5,  7:9  and  11 :  16  are  equal  respectively  to 

36  :  46,  36  :  46  and  33  :  45. 

Hence  the  ratios  4:6,  7:9  and  1 1 :  16  are  in  descending  order 
of  magnitude. 

238.  A  ratio  is  altered  in  value  when  the  same  quan- 
tity is  added  to  each  of  its  terms. 

For  example,  by  adding  1,  10  and  100  to  each  of  the  terms  of 
the  ratio  4  :  5,  we  obtain  respectively  the  ratios 

6:6,  14 :  16  and  104  :  105  ; 

and  these  new  ratios  are  different  from  the  given  ratio  and  from 
each  other. 

Since 

we  see  that  by  adding  the  same  quantity  to  each  of  the  terms  of 
the  ratios  4  :  5,  a  new  ratio  is  obtained  which  becomes  more  nearly 
equal  to  unity  as  the  quantity  added  becomes  greater. 

This  is  a  particular  case  of  the  following  general  proposition. 

239.  A)iy  ratio  is  made  more  nearly  equal  to  unity  by 
adding  the  same  positive  quantity  to  each  of  its  terms. 

By  adding  x  to  each  term  of  the  ratio  a  :  b,  the  ratio 
{a -{- x)  :  {b  -{- x)  is  obtained.  We  have  to  show  that 
(a  +  a)/(6  +  x)  is  more  nearly  equal  to  1  than  is  a/b. 


PROPORTION 

VARIATION. 

t-'- 

a  — 
b 

b 

a-\-x 

_1_ 

a- 

-b 

346  RATIO 

Now 

also  ^_     — , 

b  -j-  X  b  -i-x 

and  it  is  clear  tliat  the  absolute  value  of  (a  —  b)/(b  +  x) 
is  less  than  that  of  (a  —  b)/b,  for  the  numerators  are  the 
same  and  the  denominator  of  the  first  is  greater  than 
that  of  the  second  :  this  proves  the  proposition. 

Now  when  a  is  greater  than  b,  a/b  is  greater  than  1, 
and  so  also  is  {a-\-x)/{b-{- x)-,  hence,  as  (a -\- x)  / (b  -\-  x) 
is  more  nearly  equal  to  unity  than  a/b  is,  it  follows  that 
{a -\- x)  /  {b -{-  x)  is  less  than  a/b. 

Thus  a  ratio  which  is  greater  than  unity  is  diminished 
by  addi7ig  the  same  positive  quantity  to  each  of  its  terms. 

If,  however,  a  is  less  than  b,  a/b  is  less  than  1,  and 
so  also  is  (a  +  x)/{b  +  x);  but  (a  +  x)/{b  +  x)  is 
more  nearly  equal  to  unity  than  a/b'  is,  and  therefore 
{a  -\-x)/(b  -^  x)  is  greater  than  a/b. 

Thus  a  ratio  which  is  less  than  unity  is  increased  by 
adding  the  same  positive  quantity  to  each  of  its  terms. 

Also,  when  x  is  very  great,  the  fraction  {a  —  b)/(b-\-x) 
is  very  small ;  and  we  can  make  (a  —  b)/(b  +  x),  which 
is  the  difference  between  {a -}- x)  /  {b -\-  x)  and  1,  as  small 
as  we  please  by  taking  x  sufficiently  great.  This  is  ex- 
pressed by  saying  that  the  limit  of  (a-\-x)/(b-\-x)  when  x 
is  very  great,  is  unity. 

240,   The  following  definitions  are  sometimes  required : 
The  ratio  of  the  product  of   the  first  terms  of   any 
number  of  ratios  to  the  product  of  their  second  terms 
is  called  the  ratio  compounded  of  the  given  ratios. 


RATIO.      PROPORTION.      VARIATION.  347 

Thus  ac :  bd  is  called  the  ratio  compounded  of  the 
ratios  a  :  b  and  c:  d. 

The  ratio  a^ :  b^  is  called  the  duplicate  ratio  of  a:b. 

The  ratio  a^ :  b^  is  called  the  triplicate  ratio  of  a:b. 

The  ratio  -y/a :  -^b  is  called  the  sub-duphcata  ratio  of 
a :  b. 

241.  Incommensurable  Numbers.  The  ratio  of  two  quan- 
tities cannot  always  be  expressed  by  the  ratio  of  two 
whole  numbers  ;  for  example,  the  ratio  of  a  diagonal  to  a 
side  of  a  square  cannot  be  so  expressed,  for  this  ratio  is 
^2,  and  we  cannot  find  any  fraction  which  is  exactly 
equal  to  ■y/2. 

When  the  ratio  of  two  quantities  cannot  be  exactly 
expressed  by  the  ratio  of  two  whole  numbers,  they  are 
said  to  be  incommensurable. 

Although  the  ratio  of  two  incommensurable  numbers 
cannot  be  found  exactly,  the  ratio  can  be  found  to  any 
degree  of  approximation  which  may  be  desired ;  and  the 
diiferent  theorems  which  are  proved  with  respect  to  ratios 
of  commensurable  numbers  can  be  proved  to  be  true  also 
for  the  ratios  of  incommensurable  numbers. 

EXAMPLES  LXX. 

1.  Arrange  the  ratios  5  :  6,  7  :  8,  41  :  48,  and  31  :  36  inglescend- 
ing  order  of  magnitude. 

2.  For  what  value  of  x  will  the  ratio  3  +  a; :  4  +  x  be  equal  to 
5:6? 

3.  For  what  value  of  x  will  the  ratio  \b  -\-  x  :  11  +  x  be  equal 
to  ^? 

4.  What  must  be  added  to  each  of  the  terms  of  3  :  4  to  make 
the  ratio  equal  to  25  :  32  ? 


348  RATIO.      PROPORTION.      VARIATION. 

5.  Find  two  numbers  whose  ratio  to  one  another  is  5  :  6,  and 
whose  sum  is  121. 

6.  Two  numbers  are  in  the  ratio  3  to  8,  and  the  sum  of  their 
squares  is  3577  :  find  them. 

7.  What  is  the  ratio  of  x  to  y,  if  ^J^"^^^  =  2  ? 

Zx-y 

8.  If  4  x2  -f  ?/2  —  4  xy^  find  the  ratio  of  x  to  y. 

9.  Find  x  :  y,  having  given  x^  +  Qy^  =  5  xy. 

10.  A  certain  ratio  will  be  equal  to  2  :  3  if  2  be  added  to  each 
of  its  terms,  and  it  will  be  equal  to  1 : 2  if  1  be  subtracted  from 
each  of  its  terms  :  find  the  ratio. 

11.  Find  two  numbers  such  that  their  sum,  their  difference,  and 
the  sum  of  their  squares  are  as  7  : 1 :  76. 

12.  What  is  the  least  integer  which  must  be  added  to  the  terms 
of  the  ratio  9  :  23  in  order  to  make  it  greater  than  the  ratio  7:11? 

13.  Write  down  the  ratio  compounded  of  the  ratios  2 : 3  and 
15  :  16  ;  also  the  ratio  compounded  of  the  ratios  5  :  6  and  18  :  25. 

14.  Write  down  two  quantities  which  are  in  the  duplicate  ratio 
of  2  ic  :  3  ?/. 

15.  Find  x  in  order  that  a;  +  1 :  a;  +  4  may  be  the  duplicate 
ratio  of  3:5. 

16.  The  ages  of  two  persons  are  as  3  :  4  and  thirty  years  ago 
they  were  as  1 :  3,  what  are  their  present  ages  ? 

17.  Show  that,  if  from  each  term  of  a  ratio  the  inverse  of  the 
other  be  taken,  the  ratio  of  the  differences  will  be  equal  to  the 
original  ratio. 

18.  Show  that,  if  a  and  x  be  positive  and  a  >  x,  then  a^  —  x^: 
a2  +  x^  will  be  greater  than  a  —  x  :  a  +  x. 


RATIO.      PROPORTION.      VARIATION.  349 

PROPORTION. 

242.  Definition.  Four  quantities  are  said  to  be  propor- 
tional when  the  ratio  cf  the  first  to  the  second  is  equal  to 
the  ratio  of  the  third  to  the  fourth. 

Thus  a,  b,  c,  d  are  proportional,  if 

a:b  =  c:d. 

This  is  sometimes  expressed  by  the  notation 

a:b::c:dy 

which  is  read  "  a  is  to  6  as  c  is  to  d." 

The  first  and  fourth,  of  four  quantities  in  proportion, 
are  sometimes  called  the  extremes,  and  the  second  and 
third  of  the  quantities  are  called  the  means. 

243.  If  the  four  quantities  a,  b,  c,  d  are  proportional, 
we  have  by  definition 

b^d 
Multiply  each  of  these  equals  by  bd ;  then 
ad  =  be. 

Thus  the  product  of  the  extremes  is  equal  to  the  product 
of  the  means. 

Conversely,  if  ad  =  be  then  a,  b,  c,  d  will  be  propor- 
tional. 

For,  if  ad  =  be, 

then  qd^bc,     .    a^e^ 

bd     bd  b     d 

that  is  a:b  =  e:  d. 


350  RATIO.      PROPORTION.      VARIATION. 

Thus,  if  a:b  =  c:d,  then  ad  =  bc; 

and  conversely  if 

ad  =  be,  then  a:b  =  c:d. 

244.  It  follows  from  the  last  article  that  the  four  rela- 
tions 

a:b  =  c:d, 

a:c=b:dj 

b:a  =  d:Cf 

and  b:  d=:a:c, 

are  all  true,  provided  that  ad  =  bc. 

Hence  the  above  four  proportions  are  all  true,  when 
any  one  of  them  is  true. 

245.  If  a:b  =  c:d, 

then  will  a-\-b:  a  —  b  =  c-\-d:c  —  d. 

For  a-{-b:a  —  b  =  c-\-d:c  —  d, 

if  {a-^b)(c  —  d)  =  {a  —  b){c-\-d), 

that  is,  if   ac-\-bc  —  ad  —  bd  =  aG—bc-\-  ad  —  bd, 

or,  if  be  =  ad. 

But  this  condition  is  satisfied,  since  a  :  b  —  c  :  d. 

The  above  proposition  has  already  been  proved  in  Art. 
170,  Ex.  1. 

246.  Definitions.  Quantities  are  said  to  be  in  continued 
proportion  when  the  ratios  of  the  first  to  the  second,  of 
the  second  to  the  third,  of  the  third  to  the  fourth,  etc., 
are  all  equal. 


RATIO.      PROPORTION.      VARIATION.  351 

Thus  a,  b,  c,  d,  etc.,  are  in  continued  proportion  if 
a:6  =  6:c  =  c:d,  etc. , 

that  is,  if  ^  =  ^  =  2  =  etc. 

bed 

If  a:b  =  b  :c,  then  h  is  called  the  mean  proportional 
between  a  and  c ;  also  c  is  called  the  third  proportional  to 
a  and  b. 

247.  If  a,  b,  c  be  in  continued  proportion,  we  have 

a_b 
b~c' 
.  • .  b"^  =ac,  OT  b=  -\/ac. 
Thus  the  mean  proportional  between  two  given  quantities 
is  the  square  root  of  their  product. 

Also  ^x^  =  ^x^ 

b      0      c      b 

that  is  ^'  =  -. 

Hence  a  :  c  —  a^ :  b^. 

Thus,  if  three  quantities  are  in  continued  proportion  the 
ratio  of  the  first  to  the  third  is  the  duplicate  ratio  of  the 
first  to  the  second. 

248.  It  is  often  very  convenient  to  represent  a  ratio 
by  a  single  letter,  as  in  Art.  170.  The  following  are 
additional  examples : 

Ex.  1.   If      a'.b  =  c:d, 
then  a2  +  a5 :  c2  +  cd  =  &2  _  2  a6  :  (22  -  2  cd. 

Let  -  =  x;  then  also  -  =  x. 

b  d 


352  RATIO.      PROPORTION.      VARIATION. 

Hence  a  =  bx,  and  c  =  dx. 

(jfi  +  ah  ^  6^x2  +  h-^x  ^  h'^jpfi  +  x)  ^  6^ 
'  '  c^  +  cd  d^x'^  4-  d^x  d\x:^  +  x)  d^' 
Also  6'^  -  2  q^  ^  62  _  2  6%  ^  62(i_2x)  ^  6^^ 

d^-2c(^     #_2(i2a;     (i2(^i_2x)      d^' 

Hence  ai±^&  ^  ^!zil^, 

that  is  a2  +  a& :  c2  4-  cd  =  &2  _  2  a&  :  ^2  _  2  c(?. 

Ex.  2.    If  a:  b  =  c:d  =  e:f^  show  that 

a3  +  c3  +  e3  :  63  +  ^3  ^^-3  ^  ace  :  6d/. 

Let  a/6  =  x  ;  then  c/d  =  x  and  e/f  =  x. 

Hence  a  =  bx,  c  =  dx,  and  e=fx; 

a^  -{-  c^  -\-  e^  _  b^x^  +  d^x^  +  f^x^  _  „3 
•*•  684.^34.^-3-       534.  ^3  4.^3 

And  —  =  ^^'^^'f^  -  ic8 

bdf  bdf 

Hence  «i±ci±^  ^  ace, 

63  +  #+/3      6d/ 
that  is  a^  +  c8  +  e8 :  68  4.  (^3  +y3  _  ^qq  .  ^^f^ 

Ex.  3.   Show  that,  if 

x:26  +  2c-a  =  t/:2c  +  2a-6  =  0:2a  +  26-c, 
then  will  a:2?/4-20  —  x  =  6:22!  +  2ic  —  y  =  c:2x  +  2y-2;, 

We  have       — 


26  +  2c-a     2c  +  2a-6     2a  +  2.6-c 
Put  A  for  each  of  these  equal  fractions  ;  then 

ic  =  A(26 +  2c-a),   2/  =  A(2c  +  2a-6), 

;s  =  \(2a  +  26  -  c). 
Hence  22/4-22!  —  a;  =  9  aA,  and  similarly 

20  +  2x  —  ?/  =  9  6a  and  2x+2y  —  0  =  9cA. 
a  6  c 


Whence 


2y  +  20-x      2«  +  2x-i/      2x  +  2y-;2 


EATIO.      PROPORTION.      VARIATION.  353 

249.  The  definition  of  proportion  given  in  Euclid  is 
as  follows:  Four  quantities  are  proportional,  when  if 
any  equimultiples  be  taken  of  the  first  and  the  third,  and 
also  any  equimultiples  of  the  second  and  the  fourth,  the 
multiple  of  the  third  is  always  greater  than,  equal  to,  or 
less  than  the  multiple  of  the  fourth,  according  as  the 
multiple  of  the  first  is  greater  than,  equal  to,  or  less  than 
the  multiple  of  the  second. 

If  the  four  quantities  a,  b,  c,  d  satisfy  the  algebraical 
test  of  proportionality,  we  have 

b~d'' 

therefore,  for  all  values  of  m  and  n, 

ma  _mc 
nb      nd 
Hence 

mc>=  or  <nd,  according  as  ma  >  =  or  <  w6. 

Thus  a,  b,  c,  d  satisfy  also  Euclid's  test  of  proportionality. 

Next,  suppose  that  a,  b,  c,  d  satisfy  Euclid's  definition 
of  proportion. 

If  a  and  b  are  commensurable,  so  that  a:b  =  m:n, 
where  m  and  n  are  whole  numbers  ;  then 


a 
b^ 

m 

na  = 

=  mb. 

But,  Euclid's  definition  of  the  proportion  a:b=c:d 
asserts  that 

nc  >  =  or  <  md,  according  as  wa  >  =  or  <  mb. 

z 


354  RATIO.      PROPORTION.      VARIATION. 

Hence,  as  na  =  mb,  we  must  have 

nc  —  md ; 

c  _m  _a 
'  d      n      b 

Thus  a,  b,  c,  d  satisfy  the  algebraical  definition. 

If  a  and  b  are  incommensurable  we  cannot  find  two 
whole  numbers  m  and  n  such  that  a  :  b  =  m  :  n.  But,  if 
we  take  any  multiple  na  of  a,  this  must  lie  between  two 
consecutive  multiples,  say  mb  and  (m  + 1)6,  of  b,  so  that 

na  >  mb  and  na  <  (m  +  1)6. 

Hence,  by  the  definition, 

nc  >  md  and  nc  <  (m  -f  l)c? ; 

.*.  —  >—  and  -< — ^!^ — 
d      n  d         n 

Hence  both  a/b  and  c/d  lie  between 

™  and  '?^+l. 
n  n 

Thus  the  difference  between  a/b  and  c/d  is  less  than  1/n ; 
and  as  this  is  the  case  however  great  n  may  be,  a/b 
must  be  equal  to  c/d. 

EXAMPLES  LXXI. 
1.    Show  that,  if  a:b::c:d,  then 

(i.)  ac'.bd-.id^icP. 
(ii.)  abicd:  :a^:c^. 
(iii.)  a2:c2::a2_52.c2_<22. 

3.    Show  that,  if  a:b  =  c:d,  then 

2a  +  3c:3a  +  2c  =  26  +  3d:36+2(!. 


RATIO.      PROPORTION.      VARIATION.  355 

3.  If  a :  6  ::  a  +  c :  6  +  cZ,  then 

c:d::c  -{-  a:d  +  b. 

4.  If  a  :  b  =  c  :  d,  then 

la  +  mb:pa  +  qb  =  lc-\-  md:pc  +  qd. 

5.  Show  that,  if  3a-56:3c-5d  =  5a  +  36:5c-p3d, 
then  a:b  =  c:d. 

6.  Find  a  mean  proportional  to  a^b  and  ab^. 

7.  Find  a  mean  proportional  to  (a  -\-  by  and  (a  —  by. 

8.  Find  a  third  proportional  to  a  and  a^  ;  also  to 

(a  -  by  and  a^  -  62. 

9.  If  a:b::c:d,  then  will  ab  +  cd  he  a  mean  proportional 
between 

a2  +  c2  and  b"^  +  d^. 

10.    Show  that,  if  a:b::c:d,  then 

(i.)  a:  a  +  c:  :a-\-  b:  a  +  b  -{■  c  +  d. 
(ii.)  «2  +  0^6  4.  52 .  <j2  _  055  ^  ^2 . .  c2  +  cd  +  d-^ :  c2  -  cdJ  +  d2. 


(iii.)  a  +  6  :  c  +  d : :  Va^  +  6-^ :  VcM^. 

(iv.)  Va2  +  62  :  Vc2  +  (^^ :  :  v^a^  +  6* :  v^cS  +  d*. 

(v.)  a2c  _|.  ac2  :  62d  +  6^2 : :  (a  +  c)* :  (6  +  d)». 

(vi.)  \/a»»  +  6" :  Vc"  +  d" : :  >/a'-  -  6'' :  \/c^  -  (?•. 

11.  If  ?  =  ^  =  ?,  then  will  ^L±y  =  yjtl  =  L±J^,  and  also 

a     b     c  a+6      6+cc+a 

(a2  +  ^,2  4.  c2)(x2  4.  y2  4.  2-2)  =  (^X  +  6^  +  C«)2. 

12.  Show  that,  if 

bz  —  cy  _cx_—_az  _  ay  —  bx 
a  b  c 

then  will  ?  =  2^  =  1 

a     b     c 


356  RATIO.      PROPORTION.      VARIATION. 

13.  Show  that,  if        (^Y  +  (^)'  =  2^' 
then  will  a:h  =  c:d. 

14.  Show  that,  if 

lx{ny  —  mz),   mijilz  —  nx)   and  nz{mx  —  ly) 
be  equal  and  not  zero,  then  will 

mn  +  nl-{-  Im  =  0  and  2/0  +  0X  +  ojy  =  0. 

15.  Show  that,  if 

X     V    _         y        _        z 
h -{■  c  —  a     c  +  a  —  b     a  +  b  —  c 
then  will  (b  —  c)x  +  (c  —  a)y  +  («  —  b)z  =  0. 

16.  Show  that,  if  a:b  =  c:d  =  e:f,  then 

a^  +  a^c  +  ace  :  b^  +  b^d  +  bdf: :  ace  +  ac^  +  c^ :  6(?/+  6(^2  ^_  ^ 

17.  If  a :  a; :  :  6  : 2/ :  :  c  :  0,  prove  that 

a*  +  a--^62  +  6*  :  a;4  +  ic2?/2  +  y* :  :  6*  +  62c2  _|.  c*  :  yi  +  ?/2^2  +  ^. 

18.  Show  that,  if  a  :  b  =  pa  —  qc  :  pb  —  qd,  then  will 

c :  d  =  pa  +  qc  :  pb  +  qd. 

19.  Show  that,  if  a:b  —  c:d,  then 

bade 

20.  If  4  a  -  & :  4  a  +  6  : :  1 : 2,  find  the  value  of 

7a  +  36:7a-3  6. 

21.  If  icy  +  3  :  a;0  +  1  =  2/2  +  3  : 2/0  +  1, 
show  that  x  =  y  or  ?/  =  3  0. 

22.  Show  that,  if  a  +  6  —  c:c  +  d  +  a  =  a  —  c:2(f, 
then  b  :  a  —  c  =  a  -\-  c  —  d  -.2  d. 

23.  Show  that,  if  x  -  z:y  —  z —  x^-.y'^, 

then  X  +  2:1/  +  2r  =  a;2  +  2  xy :  2/2  +  2  a;2/. 


RATIO.      PROPORTION.      VARIATION.  357 

24.  Show  that,  if  a(y-z)+  h(z  -  x)  +  c{x  -y)  =  0, 
then  y  —  z:b  —  c  =  z  —  x:c  —  a  =  x  —  y.  a  —  b. 

25.  Show  that,  if  a:b:  :c:d, 

then  will  a^  +  b^  +  c^  +  d^ :  (a  +  by  +  (c  +  d)^  : :  (a  +  g)2^ 

+  (6  +  rf)2:(a  +  6  +  c  +  (i)2. 

VARIATION. 

250.  When  any  substance  is  sold  at  a  fixed  price  per 
pound,  the  cost  of  any  amount  of  it  is  so  related  to  its 
weight  that  when  the  weight  is  doubled  the  cost  is  also 
doubled,  when  the  weight  is  halved  the  cost  is  also 
halved,  and  so  on,  the  ratio  of  any  two  values  of  the  cost 
being  equal  to  the  ratio  of  the  corresponding  Aveights. 
When  two  quantities  are  related  in  this  way  they  are 
said  to  vary  as  one  another. 

Definition.  One  magnitude  is  said  to  vary  as  another 
when  the  two  are  so  related  that  the  ratio  of  any  two 
values  of  the  one  is  equal  to  the  ratio  of  the  correspond- 
ing values  of  the  other. 

Thus,  if  tti,  tta,  be  any  two  measures  of  one  of  the 
quantities,  and  b^,  h^j  be  the  corresponding  measures  of 
the  other,  we  have 

-^  =  -1 ;  and  therefore  -^  =  ^• 

Hence  the  measures  of  corresponding  values  of  the  two 
magnitudes  are  always  in  the  same  ratio. 

251.  The  symbol  oc  is  used  for  the  words  varies  as: 
thus  ^  oc  jB  is  read  "  A  varies  as  B." 


358  RATIO.      PROPORTION.      VARIATION. 

If  a  X  6,  the  ratio  a:  b  is  constant ;  and  if  we  put  m 
for  this  constant  ratio,  we  have 

-=m;    .'.  a  =  mb. 
b 

To  find  the  constant  m  in  any  case  it  is  only  necessary 
to  know  one  set  of  corresponding  values  of  a  and  b. 

For  example,  if  a  x  6,  and  a  is  10  when  b  is  2,  we  have 

a  =  mb;    .•.  10  =  w  x  2  ;     .'.  m  =  6. 
Hence  a  =  5  &. 

252.  Definitions.  One  quantity  is  said  to  vary  inversely 
as  another  when  the  first  varies  as  the  reciprocal  of  the 
second. 

Thus  a  varies  inversely  as  6,  if  ax-,  that  is  if  a  =  m  x  -,  or 
I,  b  b 

ab  =  m. 

One  quantity  is  said  to  vary  as  two  others  jointly,  when 
the  first  varies  as  the  product  of  the  two  others. 

Thus  a  varies  as  b  and  c  jointly  if  a  x  &c,  that  is  if  a  =  mbc. 

One  quantity  is  said  to  vary  directly  as  a  second  and 
inversely  as  a  third,  when  the  ratio  of  the  first  to  the 
product  of  the  second  and  the  reciprocal  of  the  third  is 
constant. 

Thus  a  is  said  to  vary  directly  as  b  and  inversely  as  c,  if 

a:b  X-  is  constant,  that  is  if  a  =  m-,  where  w  is  a  constant, 
c  c 

In  all  the  different  cases  of  variation  defined  above, 
the  constant  will  be  determined  when  any  one  set  of 
corresponding  values  is  given. 


RATIO.      PROPORTION.      VARIATION.  359 

For  example,  if  a  varies  as  6  and  inversely  as  c,  and  a  is  6  when 
6  is  2  and  c  is  9,  we  have 

a  =  W-,  and  therefore  6  =  w-- 
c  9 

H  .nee  w  =  27,  and  therefore  a  =  27  -• 

c 

253.  Theorem.  If  a  depends  only  on  h  and  c,  and  if 
a  varies  as  b  when  c  is  constant,  and  varies  as  c  when  h  is 
constant ;  then  when  both  h  and  c  vary,  a  will  vary  as  he. 

Let  a,  b,  c;  a',  b\  c  and  a",  6',  c'  be  three  sets  of  corre- 
sponding values. 

Then,  since  c  is  the  same  in  the  first  and  second 

ah  ... 

a=b' ('•) 

Also,  since  6'  is  the  same  in  the  second  and  third  cases, 
we  have 

a'      c 


(ii.) 


Hence,  from  (i.)  and  (ii.), 

a^      a'  _  be 
a'^^'~b^'' 

that  IS  — 77  =  -r-:» 

a"      b'c' 

which  proves  the  theorem. 

The  following  are  examples  of  the  above  proposition : 
The  cost  [O]  of  a  quantity  of  meat  varies  as  the  price  per 
pound  [P]  if  the  weight  [WT]  is  constant,  and  the  cost  varies  as 
the  weight  if  the  price  per  pound  is  constant ;  hence,  by  the* 
proposition,  when  both  the  weight  and  the  price  per  pound  change, 
the  cost  varies  as  the  product  of  the  weight  and  the  price. 


360  RATIO.      PROPORTION.      VARIATION. 

Thus,  if  C  cc  P,  when  W  is  constant, 

and  C  cc  W,  when  P  is  constant ; 

then  C  cc  PW,  when  both  P  and  TT  change. 

Again  the  area  [^]  of  a  triangle  varies  as  the  base  [J5]  when 
the  height  [iZ]  is  constant;  the  area  also  varies  as  the  height 
when  the  base  is  constant ;  hence,  when  both  the  base  and  the 
height  change,  the  area  will  vary  as  the  base  and  height  jointly. 

Thus,  if  Ace  B,  when  H  is  constant ; 

and  if  Ace  H,  when  B  is  constant ; 

then  A  qc  BH,  when  both  B  and  R  change. 

Again,  the  pressure  [P]  of  a  gas  varies  as  the  density  [i>] 
when  the  absolute  temperature  [  T  ]  is  constant ;  the  pressure  also 
varies  as  the  absolute  temperature  when  the  density  is  constant ; 
hence,  when  both  density  and  temperature  change,  the  pressure 
will  vary  as  the  product  of  the  density  and  temperature. 

Thus  PccDT. 

Ex.  1.   If  ^  oc  B,  and  if  also  Ace  C;  then  will  B  ce  C. 

For,  since  ^  oc  P,  we  have  A  =  mB,  where  m  is  some  constant. 

And,  since  Ace  C,  we  have  A  =  nC,  where  n  is  some  constant. 

Hence  B  = —C,   where   —  is  some  constant;    and  therefore 

poca 

Ex.  2.   If  O  oc  WP,  then  will  fFoc  ^. 

P 

For,  since  C  oc  WP,  we  have  C  =  m  -  WP,  where  m  is  some 
constant. 

1  r'  1  c 

Hence  W=—  —,  where  —  is  some  constant ;  therefore  Wee  — 
m  P  m  P 

Ex.  3.  The  pressure  of  a  gas  varies  jointly  as  its  density  and  its 
absolute  temperature  ;  also  when  the  density  is  1  and  the  tempera- 
ture 300,  the  pressure  is  15.  What  is  the  pressure  when  the 
density  is  3  and  the  temperature  is  320  ? 


RATIO.      PROPORTION.      VARIATION.  361 


Since 

Fee  TD, 

we  have 

P  =  mTD, 

where  m  is  some  constant. 

Also, 

by  the  conditions  of  the  problem, 

15  = 

wi  X  300  X  1 ; 

.'.  m  = 

A. 

.-.  P  = 

^V  ™. 

Hence  when  D  is  3  and  T  is  320,  we  have 

P  = 

jV  X  320  X  3 

- 

48. 

EXAMPLES   LXXII. 

1.  A  varies  as  B,  and  ^  is  5  when  jB  is  3 ;  what  is  A  when 
Bis  5? 

2.  W  varies  inversely  as  P,  and  TF  is  4  when  P  is  15 ;  what 
is  W  when  P  is  12  ? 

3.  U  xccy  and  y  cc  z,  then  will  xz  qc  y"^. 

4.  If  x^  X  y  and  z"^ «  y,  then  will  xz  qc  y. 

6.    If  a;  cc  -  and  y  cc  -■,  then  will  xcc  z. 
y         "     z 

6.  A  varies  as  B  and  O  jointly ;  also  ^  =  4  when  P  =  2  and 
0  =  6:  find  the  value  of  A  when  P  =  2  and  0  =  9. 

7.  A  varies  as  P  and  inversely  as  C;  also  A  =  2  when  P  =  3 
and  0  =  4:  find  the  value  of  P  when  ^  =  6  and  0  =  3. 

8.  The  area  of  a  circle  varies  as  the  square  of  its  radius,  and 
the  area  of  a  circle  whose  radius  is  10  feet  is  314.159  square  feet. 
What  is  the  area  of  a  circle  whose  radius  is  12  feet  ? 

9.  The  volume  of  a  sphere  varies  as  the  cube  of  its  radius,  and 
the  volume  of  a  sphere  whose  radius  is  1  foot  is  4.188  cubic  feet. 
What  is  the  volume  of  a  sphere  whose  radius  is  3  feet  ? 


362  RATIO.      PROPORTION.      VARIATION. 

10.  The  velocity  of  a  falling  body  varies  as  the  time  during 
which  it  has  fallen  from  rest,  and  the  velocity  at  the  end  of  two 
seconds  is  64.     What  is  the  velocity  at  the  end  of  five  seconds  ? 

11.  The  distance  through  which  a  heavy  body  falls  from  rest 
varies  as  the  square  of  the  time  it  falls,  and  a  bjdy  falls  through 
144  feet  in  three  seconds.     How  far  does  it  fall  in  two  seconds  ? 

12.  Given  that  the  area  of  a  circle  varies  as  the  square  of  its 
radius,  show  that  a  circle  of  5  feet  radii:is  is  equal  to  the  sum  of  a 
circle  of  3  feet  radius  and  another  of  4  feet  radius. 

13.  The  volume  of  a  gas  varies  as  the  absolute  temperature  and 
inversely  as  the  pressure.  Also  when  the  pressure  is  15  and  the 
temperature  280  the  volume  is  1  cubic  foot ;  what  is  the  volume 
when  the  pressure  is  20  and  the  temperature  300  ? 

14.  If  d?'  —  62  varies  as  c^,  and  if  c  =  2  when  a  —  h  and  6  =  3: 
find  the  equation  between  a,  6,  and  c,  and  show  that  6  is  a  mean 
proportional  between  a  —  2  c  and  a  +  2  c. 

15.  The  volume  of  a  right  circular  cone  varies  jointly  as  its 
height  and  the  square  of  the  radius  of  its  base  ;  and  the  volume  of 
a  cone  7  feet  high  with  a  base  whose  radius  is  3  feet  is  66  cubic 
feet.  Find  the  volume  of  a  cone  9  feet  high  with  a  base  whose 
radius  is  14  feet. 

16.  The  volume  of  a  sphere  varies  as  the  cube  of  its  radius  ;  if 
three  spheres  of  radii  6,  8,  and  10  inches  respectively  be  melted  and 
formed  into  a  single  sphere,  find  its  radius. 

17.  The  distance  of  the  offing  at  sea  varies  as  the  square  root  of 
the  height  of  the  eye  above  the  sea  level,  and  the  distance  is  3  miles 
when  the  height  is  6  feet ;  find  the  distance  when  the  height  is  50 
yards. 


MISCELLANEOUS   EXAMPLES   V.  363 

MISCELLANEOUS  EXAMPLES      V. 

A.  1.   If  —  =  —  =  -  =  1,  find  the  numerical  value  of 

3        4       2 

(a  -  6)2  +  (6  -  c)2  +  (c  -  a)2. 

2.  Divide 

9  a263  _  I2a46  +  3  65  +  2  0^62  +  4  a^  _  n  «54  by  3  63  +  4  a^  -  2  a62. 

3.  Find  the  H.  C.  F.  of  1  -  x  -  x^  +  x^  and  1  -  ic^  _  a^  _  ^7. 

4.  Simplify  a^-a^b-^-l  +  b-^. 

^    ^    a  +  a6-i  + 1  +  6-1 

6.  A  person  bought  15  ducks  and  12  geese  for  $26.25  ;  and  the 
prices  were  such  that  2  more  ducks  could  be  bought  for  $4.50,  than 
geese  for  $5.     What  was  the  price  of  a  duck  ? 

6.  Show  that  the  difference  of  the  roots  of  the  equation 
x^  —  px  -\-  q  =  0  is  equal  to  the  difference  of  the  roots  of  the  equa- 
tion x2  -  Spx  +  2i)2  +  g  =  0. 

7.  What  is  the  least  integer  which  must  be  added  to  the  terms 
of  the  ratio  3  :  4  to  make  it  greater  than  the  ratio  19  :  21  ? 

8.  Show  that,  if  a  +  6,  6  +  c,  c  +  a  are  in  continued  proportion, 
then  b  +  c  :  c  +  a  :  :  c  —  a  :  a  —  b. 

B.  1.   Find  the  value  of 

(Xy  +  Vl  -  X2V1  -  2/2)  ^  (xVl  -  !/2  _  yy/l  _  a;2), 

when  X  =  I,  y  =  |. 

2.  Multiply  x2  +  (a  -  l)x  +  a  +  1  by  (a  -  l)x  -  a2  -  a  -  1. 

3.  Find  the  factors  of 

(i.)  x8-  13x2^4- 42  xy2, 
(ii.)  (a  +  2  6  +  3  c)2  -  4(a  +  6  -  c)2, 
and  (iii.)  x2  —  4  xy  +  4  2/2  _  9. 

4.  Simplify        ^      +  -^^-  +      ^^^     ♦ 

^    ^  x-2y     x-\-2y     iy^-x^ 


364  MISCELLANEOUS   EXAMPLES  V. 

6.    Solve  the  equations  : 

,j.  10 3     ^    10 

^        X      x-\-2     x  +  1 
(ii.)  x  +  y4-v'(^  +  y)=12l 

iK2  -  2/2  _  21  j   ' 

6.  If  a;  +  -  =  1  and  ?/  +  -  =  1,  prove  that  z  +  -  =  1.  and  that 

y  "      z  "^  X 

xyz  +  l=  0. 

7.  Find  the  square  root  of 

9  a;6  -  12  x'^y'^  +  30  x^y^  +  4  x^y^  -  20  xy^  +  25  ?/«. 

8.  Find  two  numbers  such  that  their  sum,  their  difference,  and 
the  sum  of  their  squares  are  as  3  : 1 :  15. 

9.  It  a:b  :  :  c:d,  show  that 


la+mb:lc+  md:  :  Va^  +  b'^d  :  Vac"^  +  #. 
C.     1.    Subtract  a  -  2(6  -  c)  from  3  {&  +  2(a  -  c)  -  5(a  -  6)}. 

2.  Multiply  a  +  ?)  +  -  +  —  bya-6  +  ^-^• 

a       6     *  a       b 

3.  Multiply     ^'^"^^^     by  27 (a  -  &),  and  divide 

9(a2  _  ^2) 

9a2_i6  62        Sa-4tb 
a+b         ^    a^-b^' 

4.  If  a  =  y  +  0,  b  =  z  -{-  X,  c  =  jc  +  y,  then 

a2  +  62  _|.  c2  _  6c  —  ca  —  a6  =  a;2  _i_  ^2  ^.  ;j;2  _  2/;2!  _  5!X  —  a;?/. 

5.  Find  the  square  root  of 

aJ  +  4  ay^  +  10  a^y^  +  12  a^y^  +  9  y^. 

6.  If  x  =  2  +  V2,  prove  that  (x  -l)(x-2)  =  x.     . 

7.  ^  and  5  start  simultaneously  from  two  towns  to  meet  each 
other  ;  A  travels  two  miles  an  hour  slower  than  B,  and  they  meet 
in  seven  hours  ;  if  B  had  travelled  one  mile  an  hour  faster  than  he 
did  and  A  at  only  half  his  previous  pace,  they  would  have  met  in 
nine  hours.    Find  the  distance  between  the  towns. 


MISCELLANEOUS   EXAMPLES  V.  865 

8.   If  a(y  +  z)=  b(z  +  x)=  c(x+  y)  ;  then 
y  —  z    _    z  —  X    _    x  —  y 
a{h  —  c)      h(c-a)      c(a  —  b) 

D.    1.   Show  that  («i^  V  -  ( ^+^Y  =  4  aKa^  +  b^). 
\  a  —  b  J       \a  -\-  b  J 

2.   Divide 

a4  -  2  6x8  -  ((j2  _  52)ic2  +  2  a^bx  -  a'^b^  by  x^  -  (a  +  b)x  +  ab. 

8.  Find  the  L. CM.  of 

x2-7x+  12,  3x2 -6x- 9  and  2x8- 6x2 -8x. 


X2;     2:y  +  z^ 
5.   Solve  the  equations : 


5x  +  7_2x-J^3        14^ 
^  ^        2  3 

(ii.)  y2_a;y  =  161 

x2  +  a;y  =  14  i 

6.  The  fore-wheel  of  a  carriage  makes  64  revolutions  more  than 
the  hind-wheel  in  travelling  one  mile,  and  the  sum  of  the  revolu- 
tions made  by  each  in  a  journey  of  10  miles  is  7040  :  find  the  cir- 
cumference of  each  wheel. 

7.  Divide  2  x^  -  17  xy^  +  17  x^y  -  15  y^  by  2  x^  -  15  y^. 

8.  Prove  that  if  the  ratios  a:  a,b  :  ^^  and  c :  y  are  all  equal, 
then  each  is  equal  to  the  ratio  a-|-64-c:o-|-/34-7. 

E.   1.    Simplify  the  expression 
2{x-2a  -|(x-  a)+  a]  -  l{(x  -  a)-(a  -  b)  +  y  -b}. 

2.  Divide  2  a2x2  -  2(3  6  -  4c) (&  -  c) 2/2 -|-  a6xy  by  ax+2(6-c)y. 

3.  Pind  the  factors  of  the  following  expressions  : 

(i.)  4a26*-16a*62,  (ii.)  x2-26x-87,  and  (iii.)  'Sx^-xy-lOyK 


866  MISCELLANEOUS   EXAMPLES   V. 

4.  Show  that 

1      +      1      +      1      =f_!_  +  _L-  +  -J_Y. 

(y  -  z)'^      iz-x)^     (x-yY     \y  -  z     z-x     x-y) 

5.  A  man  bought  a  number  of  articles  for  $  108.  If  he  had  got 
five  more  for  the  same  money  each  would  have  cost  24  cents  less 
than  it  did  ;  how  many  did  he  buy  ? 

6.  Multiply  y/(x  -  a)  +  y/a  —  y/x'by  V(^  -  «)  -  V«  +  ^x. 

7.  Show  that  the  mean  proportional  between  x"^ and  y^ 

is  xy 

xy 

8.  li  a:h'.:c:d,  show  that 

a2  +  62  :  c2  +  d:2  .  .  (a  +  5)2  .  (c  +  ay, 

F.    1.    Multiply  a  —  h  +  c  —  d\iYa-\-h  —  c-\-d. 

2.  Divide  a^  +  63  +  c^  +  3(6  +  c)  (c  +  a)  (a  +  6)  by  a  +  6  +  c. 

3.  Find  the  H.  C.  F.  of  x^  -  19  x  +  30  and  5  x^  -  19  a;2  +  36.  For 
what  values  of  x  will  both  expressions  vanish  ? 

4.  Show  that  /a^&y_/a:^Y^8a6(a2  +  62). 

6.  Solve  the  equations  : 

^'^  9x  +  6      12x4-8      2* 

(ii.)  4x  +  62/=    3| 

4x2  4-9x?/  +  9y2=^  11  j  * 

_2 

and  x^y^  x  (  —  W  {x"*y"^}. 

7.  Find  the  square  root  of 

a;i2  _  6x10  +  13x8  -  14x6  +  lOx*  -  4x2  +  i. 

8.  If  a  :  6  :  :  c  :  d,  show  that  aH^  :  63^2  :  :  a^  +  c^  :  ft^  +  d^. 


ABITHMETICAL  PROGRESSION.  367 


CHAPTER   XXIV. 
Arithmetical  Progression. 

254.  Def.  A  series  of  quantities  is  said  to  be  in  Arith- 
metical Progression  when  the  difference  between  any  term 
and  the  preceding  one  is  the  same  throughout  the  series. 

Thus  a,  h,  €,  d,  etc.,  are  in  Arithmetical  Progression 
(A.P.)  if  b  —  a=zc  —  b  =  d—c,  etc. 

The  difference  between  each  term  of  an  AP.  and  the 
preceding  term  is  called  the  common  difference. 

The  following  are  examples  of  arithmetical  progressions : 


1, 

3, 

6, 

7,  etc., 

2, 

6, 

10, 

14,  etc.,                ^ 

9, 

8, 

7, 

6,  etc.,               -^ 

and                             3,     - 

-1, 

-5, 

-9,  etc. 

In  the  first  series  the 

common 

difference  is  2,  in  the  second 

series  it  is  4,  in  the  third  it  is 

—  1,  and  in  the  last  it  is  —4. 

255.   If  the  first  term  of  an  arithmetical  progression 
be  a,  and  the  common  difference  d,  then 

the  2d    term  will  he  a  +  d, 
the  3d    term  will  be  a  4-  2  c^, 
the  4th  term  will  he  a -\- 3d, 

and  so  on,  the  coefficient  of  d  being  always  less  by  unity 
than  the  number  giving  the  position  of  the  term  in  the 
series. 


368  ARITHMETICAL   PROGRESSION. 

Thus  the  nth  term  is 

a-\-{n  —  l)d. 

We  can  therefore  write  down  any  term  of  an  A.P. 
when  the  first  term  and  the  common  difference  are  given. 

For  example,  in  the  A.P.  whose  first  term  is  5  and  whose  com- 
mon difference  is  3, 

the  9th  term  is  5  +   (9  -  1)3  =  29, 
and  the  27th  term  is  5  +  (27  -  1)3  =  83. 

256.  When  any  two  terms  of  an  A.P.  are  given,  we 
can  find  the  first  term  and  the  common  difference,  and 
therefore  any  other  term  of  the  series. 

Suppose,  for  example,  that  the  10th  term  of  an  A.P.  is  25  and 
the  15th  term  is  5. 

Let  the  first  term  he  a  and  the  common  difference  d. 

Then  the  10th  term  will  be  a  +  9  d  and  the  16th  term  will  be 
a  +  14  d. 

Hence  a  +  9  d  =  25, 

and  a  +  14  d  =  5. 

By  subtraction  5d  =  —  20; 

.-.  d  =  -4. 

Then  a  =i  25  -  9d  =  25  -  9(- 4)=  6L 

Thus  the  series  is  61,  57,  53,  etc. 

Ex.  The  12th  term  of  an  A.P.  is  15  and  the  19th  term  is  36 ; 
find  the  30th  term. 

Let  the  first  term  be  a  and  the  common  difference  d. 
Then  the  12th  term  will  he  a  +  lid  and  the  19th  term  will  be 
a-f  18d. 

Hence,  by  the  conditions  of  the  problem, 
a  +  11  ^  =  15, 
a  +  18d  =  36. 


ARITHMETICAL  PROGRESSION.  369 

By  subtraction  7d  =  21; 

.-.  d  =  S. 
Then  a  =  15  -  11  c2  =  15  -  33  =  -  18. 

Hence  the  30th  term        =  a  +  29  (Z  =  -  18  +  29  x  3  =  69. 

257.  When  three  quantities  are  in  arithmetical  pro- 
gression, the  middle  one  is  called  the  Arithmetic  Mean  of 
the  other  two. 

Thus,  if  a,  b,  c  are  in  A. P.,  6  is  the  arithmetic  mean  of 
a  and  c. 
By  the  definition  of  arithmetical  progression,  we  have 

b  —a  =  c  —  b', 

Thus  the  arithmetic  mean  of  any  two  quantities  is  half 
their  sum. 

258.  When  any  number  of  quantities  are  in  arith- 
metical progression,  all  the  intermediate  terms  may  be 
called  arithmetic  means  of  the  two  extreme  terms. 

Between  any  two  given  quantities  any  number  of 
arithmetic  means  may  be  inserted. 

For  example,  to  msert  four  arithmetic  means  between  10  and  25. 
We  have  to  find  an  A.P.  with  four  terms  between  10  and  25,  so 
that  10  is  the  first  and  25  is  the  sixth  term  of  an  A.P. 
Let  d  be  the  common  difference. 
Then  the  6th  term  is  10  -f-  5  d ; 

.-.  10-f-5d  =  26; 
.-.  d  =  S. 
Thus  the  series  is 

10,        13,        16,        19,        22,        25; 
and  the  required  arithmetic  means  between  10  and  25  are 

13,        16,        19,        22. 
2a 


370  ARITHMETICAL   PROGRESSION. 

We  now  consider  the  most  general  case,  namely  to 
insert  n  arithmetic  means  between  a  and  b. 

We  have  to  find  an  A. P.  with  n  terms  between  a  and 
&,  so  that  a  is  the  first  and  h  is  the  (n  +  2)th  term  of 
the  A.P. 

Let  d  be  the  common  difference. 

Then  the  {n  +  2)th  term  is  a  +  (n  +  l)c^; 

.'.   {n-\-l)d=b  —  af 

.-.  d  =  ^-^^. 
n-\-l 

Thus  the  series  is 

,  b  —a         ,   nb  —  a  ,      6  —  a, 

n-\-l  n-\-l  71  +  1 

and  the  arithmetic  means  required  are 

,  b  —  a        .  (yb  —a        ,  r.b  —  a  ,     b  —  a 

n-t-l  w  +  l  n  +  1  n-\-l 

that  is, 

7ia-{-b     (n  —  l)a  +  2b     (n  —  2)a-^3b  ^^^     a -hub 

n -j- 1  n  +  1  n  +  1         '      '     n  +  l 

EXAMPLES    LXXIII. 

1.  Find  the  30th  terms  of  each  of  the  following  arithmetical 
progressions : 

(i.)  3,  5,  7,  etc.  (ii.)  1,  5,  9,  etc.  (iii.)  12,  9,  6,  etc. 

(iv.)  i,  A,  I,  etc.  (v.)  a  -{-  b,  a,  a  —  b,  etc. 

2.  Find  the  last  term  of  each  of  the  following  series : 

(i.)  3,  6,  9,  etc.,  to  24  terms.  (iv.)  14,  46,  78,  etc.,  to  12  terms, 
(ii.)  5,  9,  13,  etc.,  to  30  terms,  (v.)  6,  8f,  11^,  etc.,  to  14  terms. 
(iii.)  6,  5,  4,  etc.,  to  10  terms.      (vi.)  i,  -^,  -|,  etc,  to 25  terms. 


AEITHMETICAL   PEOGKESSION.  371 

3.  The  10th  term  of  an  A.P.  is  6  and  the  6th  term  is  10.     Find 
the  first  term. 

4.  The  12th  term  of  an  A.P.  is  15  and  the  20th  term  is  25. 
Find  the  common  difference. 

5.  The  7th  term  of  au  A.P.  is  5  and  the  12th  term  is  30.     Find 
the  common  difference. 

6.  The  3d  term  of  an  A.P.  is  40  and  the  13th  term  is  25.     Find 
the  first  term. 

7.  What  is  the  10th  term  of  the  A.P.  whose  first  term  is  7  and 
whose  third  term  is  13  ? 

8.  What  is  the  12th  term  of  the  A.P.  whose  first  term  is  20  and 
whose  6th  term  is  10  ? 

9.  The  3d  term  of  an  A.P.  is  10  and  the  14th  term  is  54.    Find 
the  20th  term. 

10.  The  7th  term  of  an  A.P.  is  5  and  the  6th  term  is  7.     What 
is  the  12th  term  ? 

11.  Which  term  of  the  series  5,  8,  11,  etc.,  is  65  ? 

12.  Which  term  of  the  series  |,  |,  |,  etc.,  is  18  ? 

13.  Which  term  of  the  series  9,  13,  17,  etc.,  is  229  ? 

14.  Which  term  of  the  A.P.  16a -8  6,  15a -7  6,  14a -66, 
etc.,  is  8a? 

16.   Write  down  the  arithmetic  mean  of  (i.)  7  and  13 ;  (ii.)  9 
and  —9,  and  (iii.)  a  +  6  and  a  —  h. 

16.  Insert  6  arithmetic  means  between  8  and  29. 

17.  Insert  8  arithmetic  means  between  50  and  80. 

18.  Insert  7  arithmetic  means  betweeen  269  and  295. 

19.  Insert  15  arithmetic  means  between  67  and  43. 

20.  Insert  25  arithmetic  means  between  84  and  40f . 

21.  Insert  10  arithmetic  means  between  5  a  —  6  6  and  5  6  —  6  a. 


372  ARITHMETICAL   PROGRESSION. 

22.  Insert  8  arithmetic  means  between  a  —  6b  and  b  —  5a. 

23.  K  a,  b,  c,  d  are  in  A.P.,  show  that  a  +  d  =  b  +  c. 

24.  The  sum  of  the  1st  and  4th  terms  of  an  A.  P.  is  19,  anw  the 
sum  of  the  3d  and  6th  terms  is  31.    What  is  the  first  term  ? 

25.  The  sum  of  the  2d  and  5th  terms  of  an  A. P.  is  32,  and  the 
sum  of  the  3d  and  8th  terms  is  48.     What  is  the  first  term  ? 

26.  The  sum  of  the  3d  and  4th  terms  of  an  A.P.  is  187,  and 
the  sum  of  the  7th  and  8th  terms  is  147.    What  is  the  second  term  ? 

27.  The  sum  of  the  2d  and  20th  terms  of  an  A.P.  is  2,  and  the 
sum  of  the  9th  and  15th  terms  is  8.  What  is  the  sum  of  the  6th 
and  7th  terms  ? 

28.  Show  that,  if  the  same  quantity  be  added  to  every  term  of 
an  A.P.,  the  sums  will  be  in  A.P. 

29.  Show  that,  if  every  term  of  an  A.P.  be  multiplied  by  the 
same  quantity,  the  products  will  be  in  A.P. 

30.  Show  that,  if  every  alternate  term  of  an  A.P.  be  taken 
away,  the  remaining  terms  will  be  in  A.P. 

31.  Show  that,  if  between  every  two  consecutive  terms  of  an 
A.P.  their  arithmetic  mean  be  inserted,  the  whole  will  form  another 
arithmetical  progression. 

32.  Show  that,  if  four  quantities  are  in  A.P.,  the  product  of 
the  1st  and  4th  is  always  less  than  the  product  of  the  2d  and  3d. 

259.  To  find  the  sum  of  any  number  of  terms  of  an 
arithmetical  progression. 

Let  a  be  the  first  term  and  d  the  common  difference. 
Let  n  be  the  number  of  the  terms  whose  sum  is  required, 
and  let  I  be  the  last  term. 


ARITHMETICAL   PROGRESSION.  373 

Then,  since  I  is  the  7ith  term,  we  have 

l=za  +  {n  —  l)d. 
Hence,  if  ;S^  be  the  required  sum, 

,S  =  a  +  (a  +  d)  +  (a  +  2  d)  +  ...  +  (Z  -  2  d)  +  (Z  -  d)  +  ?. 

Now  write  the  series  in  the  reverse  order ;  then 
^  =  Z  +  (Z  -  d)  +  (^  -  2  d)  +  ...  +  (a  -h  2  d)  +  (a  +  d)  4-  a. 
Hence,  by  addition  of  corresponding  terms,  we  have 

2>S  =  (a4-0  +  (a4-  I) -\- {a -\- 1) -\ to  w  terms 

=zn{a-Ji-l)'^ 

.-.  ^  =  ^(«  +  0 (i-) 

But    a  +  Z  =  a4-Sa  +  (w  — l)d|  =  2a  +  (n  — l)d; 

.-.  S  =  '^\2a  +  {n-l)d\ (ii.) 

The  formulae  (i.)  and  (ii.)  are  both  important  and 
should  be  remembered. 

It  should  be  remarked  that  the  formula  (ii.)  gives  the 
value  of  any  one  of  the  four  quantities  a,  d,  n,  and  S 
when  the  other  three  are  known. 

Ex.  1.   Find  the  siun  of  the  first  20  terms  of  the  series 

6  +  8  +  11  +  etc. 
Here  a  =  6,  d  =  3,  n  =  20; 

.-.  ^  =  |{2a+(n-l)d} 
=  670. 


374  ARITHMETICAL   PROGRESSION. 

Ex.  2.   Show  that  the  sum  of  any  number  of  consecutive  odd 
numbers,  beginning  with  unity,  is  a  square  number. 
The  series  of  odd  numbers  is 

1  +  3  +  5  +  7  +  ... 
Here  a  =  1  and  d  =  2;  hence  the  sum  of  n  terms  is  given  by 

S  =  ^{2a  +  {n  -  l)d} 
■       =|{2+(n-l)2} 


Thus  the  sum  of  n  consecutive  odd  numbers  beginning  with 
unity  is  n^.     [Compare  Art.  146.] 

Ex.  3.   The  sum  of  20  terms  of  an  arithmetical  progression  is 
410,  and  the  first  term  is  30.     What  is  the  common  difference  ? 

We  have  '       /S  =  ^{2a +(ri  -  1)(^}, 

where  8  -  410,  a  =  30,  and  n  =  20. 

Hence  410  =  ?^{60  +  19d}. 

From  which  we  find  that  d  =  —  1. 

Ex.  4.    How  many  terms  of  the  series  11  +  12  +  13  +  etc.  must 
be  taken  in  order  that  the  sum  may  be  410  ? 

We  have  S  =  '^{2a +(n  -  l)d}, 

where  a  =  11,  d  =  1,  and  aS'  =  410. 

Hence  410  =  ^{22 +(«  -  1)}; 

2 

.-.  w2  +  21  w- 820  =  0, 

that  is  (w-20)(w  +  41)  =  0; 

.'.  n  =  20,  or  n  =  — 41. 


ARITHMETICAL  PROGRESSION.  375 

When  a,  S,  and  d  are  given,  n  is  to  be  found  by  solving  a  quad- 
ratic equation,  and  one  of  the  roots  of  the  equation  is  generally 
inapplicable,  for  any  value  of  n  which  is  not  a  positive  integer  is 
without  meaning. 

In  the  present  case,  —  41  is  to  be  rejected  :  thus  the  only  value 
of  n  is  20. 

Ex.  6.  How  many  terms  of  the  series  24  +  21  +  18  +  etc.  must 
be  taken  in  order  that  the  sum  may  be  105  ? 


We  have 

S  =  ^{2a+in-l)d}, 

where 

a  =  24,  d  =  -  3,  and  S=  105. 

Hence 

105  =  ^{48  +  (n-l)(-3)}; 

.-.  210  =  n{48-3n  +  3}; 

...  w2_i7^  +  70  =  0. 

Hence 

n  =  7,  or  n  =  10. 

Since  both  values  of  n  are  positive  integers,  they  are  both 
answers  to  the  question.  Whenever  the  sum  of  an  A.  P.  is  the 
same  for  two  values  of  n,  the  additional  terms  for  the  greater  value 
have  a  zero  sum  ;  in  the  present  case  these  additional  terms  are  3, 
0,  -3. 

EXAMPLES   LXXIV. 

Find  the  sum  of  the  following  series : 

1.  2  +  4  +  6  +  .••  to  20  terms. 

2.  15  +  14 J  +  14  +  ...  to  16  terms. 

3.  1+2^  +  3^ ...  to  12  terms. 

4.  _  5- 1  +  3  + ...  to20terms. 
6.    i  +  i  +  i  H to  7  terms. 

6.  i  -  §  -  Y  -  •••  to  61  terms. 

7.  10  +  ^^-1- ^/+ •••  to  7  terms. 


376  ARITHMETICAL   PROGRESSION. 

8.  1  + 1  +  f -f  •..  to  15  terms. 

9.  3i  +  2^  +  1|  +  •••  to  n  terms. 

10.  1^  +  IH-  +  Hj  +  —  ton  terms. 

11.  8  +  7^  +  6f  +  ••••  to  19  terms. 

12.  4J  +  4^  +  3j9^  +  •..  to  31  terms. 

13.  5  +  6  .  2  +  7  •  4  H to  21  terms. 

14.  — = 1-  a/2  +  — h  •••  to  7  terms. 

V2  +  1      ^    ^V2-l 

15.  ^Jui  +  Vl=1  +  ?Lzi3  +  ...  to  n  terms.         - 

n  n  71 

16.  w-f  1 +(2w  +  3)  +  (3w  + 5)+ •••  to  n  terms. 

17.  (a  +  6)2+  (a2  +  62)  +  (a  -  6)2+  ...  to  n  terms. 

18.  The  3d  term  of  an  A.P.  is  15,  and  the  20th  term  is  23^ ; 
find  the  sum  of  the  first  20  terms. 

19.  The  5th  term  of  an  A.P.  is  37,  and  the  13th  term  is  81 ;  find 
the  sum  of  the  first  24  terms. 

20.  Find  the  sum  of  20  consecutive  odd  numbers  of  which  the 
least  is  25. 

21.  Find  the  sum  of  40  consecutive  odd  numbers  of  which  the 
greatest  is  99. 

22.  Insert  29  arithmetic  means  between  5  and  50,  and  find 
their  sum. 

23.  Insert  40  arithmetic  means  between  10  and  100,  and  find 
their  sum. 

24.  There  are  27  terms  of  an  A. P.,  of  which  3  is  the  first  and 
107  is  the  last.     Find  the  middle  term  and  the  sum  of  all  the  terms. 

25.  There  are  71  terms  of  an  A. P.,  of  which  the  first  is  7  and 
the  last  is  1015.  Find  the  middle  term  and  the  sum  of  all  the 
terms. 


ARITHMETICAL   PROGRESSION.  377 

26.  There  are  19  terms  in  an  A. P.,  of  which  the  first  is  8  and 
the  last  —  4.     Find  the  sum  of  the  series. 

27.  Find  the  sum  of  20  terras  of  the  series  3,  5,  7,  etc.,  beginning 
at  the  7th. 

28.  Find  the  sum  of  35  terms  of  the  series  6,  9,  12,  etc.,  begin- 
ning at  the  5th. 

29.  The  sum  of  10  terms  of  an  A. P.,  whose  first  term  is  2,  is 
155.     What  is  the  common  difference  ? 

30.  The  sum  of  25  terms  of  an  A. P.,  whose  first  term  is  6,  is 
25.     What  is  the  common  difference  ? 

31.  The  sum  of  10  terms  of  an  A.P.  is  100,  and  the  6th  term 
is  11.     What  is  the  first  term  ? 

32.  The  sum  of  28  terms  of  an  A.P.  is  133,  and  the  5th  term 
is  0.     What  is  the  common  difference  ?  ' 

33.  The  sum  of  10  consecutive  terms  of  the  series  3,  8,  13,  etc., 
is  705.     Which  is  the  first  of  them  ? 

34.  The  sum  of  25  successive  terms  of  the  series  5,  8,  11,  •••,  is 
1025.    Which  is  the  first  of  them  ? 

35.  How  many  terms  of  the  series  |,  1,  |,  must  be  taken  in 
order  that  the  sum  may  be  zero  ? 

36.  How  many  terms  of  the  series  15,  12,  9,  •••,  must  be  taken 
in  order  that  the  sum  may  be  45  ? 

1  q  C 

37.  How  many  terms  of  the  series  1 — ,    1 — ,    1—-,  •••, 

a  a  a 

must  be  taken  in  order  that  the  sum  may  be  —6a? 

38.  How    many   terms    of    the    series    —  8  —  7  —  6  —  •••  will 
amount  to  42  ? 

39.  The  first  term  of  a  certain  A.P.  is  1,  the  last  is  99,  and  the 
sum  of  all  the  terms  is  450.     How  many  terms  are  there  ? 

40.  The  last  term  of  an  A.P.  of  20  terms  is  62,  and  the  sum  is 
670.    What  is  the  common  difference  ? 


378  ARITHMETICAL   PROGRESSION. 

41.  The  ninth  term  of  an  A.  P.  is  136,  and  the  sum  of  the  first 
19  terms  is  2527.     Find  the  sum  of  the  first  40  terms. 

42.  Find  the  sum  of  15  terms  of  an  A.  P.  of  which  the  eighth 
is  6. 

43.  Find  the  sum  of  35  terms  of  an  A.P.  of  which  the  eighteenth 
term  is  15. 

44.  Find  the  sum  of  2  n  +  1  terms  of  an  A.P.  whose  {ti  +  l)th 
term  is  100. 

45.  Show  that,  if  any  odd  number  of  quantities  are  in  A.P.,  the 
first,  the  middle,  and  the  last  are  also  in  A.P. 

46.  Show  that,  if  unity  be  added  to  the  sum  of  any  number  of 
terms  of  the  series  8,  16,  24,  etc.,  the  result  will  be  the  square  of 
an  odd  number. 

•  47.    Find  the  sum  of  all  the  odd  numbers  between  100  and  200. 

48.  Find  the  sum  of  all  the  even  numbers  which  are  between 
101  and  999. 

49.  Find  the  sum  of  all  the  numbers  between  100  and  500  which 
are  divisible  by  3. 

60.  A  man  saved  each  year  $  10  more  than  he  did  in  the  pre- 
ceding year,  and  he  saved  $  20  the  first  year.  In  how  many  years 
will  his  savings  amount  to  $  1700  ? 

61.  The  sum  of  10  terms  of  an  arithmetical  series  is  145,  and 
the  sum  of  its  fourth  and  ninth  terms  is  five  times  the  third  term. 
Determine  the  series. 

62.  In  an  A.P.  consisting  of  2  n  +  1  terms,  show  that  the  sum 
of  the  odd  terms  is  to  the  sum  of  the  even  terms  a,s  n  +  1 :  n. 

63.  In  an  A.P.  consisting  of  21  terms,  the  sum  of  the  three  last 
terms  is  117,  and  the  sum  of  the  three  middle  terms  is  90.  Find 
the  series. 

64.  The  sum  of  n  terms  of  the  series  87,  85,  83,  etc.,  is  equal  to 
the  sum  of  n  terms  of  the  series  3,  5,  7,  etc.    Find  n. 


ARITHMETICAL  PROGRESSION.  379 

'    55.    The  sum  of  n  terms  of  the  A. P.  43,  45,  47,  etc.,  is  equal  to 
the  sum  of  2  w  terms  of  the  A.P,  45,  43,  41,  etc.    Find  n. 

56.  Divide  80  into  four  parts  which  are  in  A.P.,  and  which  are 
such  that  the  product  of  the  first  and  fourth  is  two-thirds  of  the 
product  of  the  second  and  third. 

57.  Find  four  numbers  in  A.P.  such  that  the  sum  of  their  squares 
shall  be  120,  and  that  the  product  of  the  first  and  last  shall  be  less 
than  the  product  of  the  other  two  by  8. 

58.  If    -,    i,    -   be  in  A.P.,   prove  that   (s  -  a)2,    (s  -  6)2, 

a     h     c 

(s  —  c)2  are  also  in  A.P.,  where  2s  =  a  +  6  +  c. 

59.  Show  that,  if  a,  6,  c  be  in  A.P.,  then  will  a\h  +  c), 
62 (c  +  a),  c2(a  +  h)  also  be  in  A.P. 


380  GEOMETRICAL  PROGRESSION. 


CHAPTER   XXV. 
Geometrical  Progression. 

260.  Def.  A  series  of  quantities  is  said  to  be  in  G-eo- 
metrical  Progression  when  the  ratio  of  any  term  to  the  pre- 
ceding one  is  the  same  throughout  the  series. 

Thus  a,  b,  c,  d,  etc.,  are  in  Geometrical  Progression 

(G.P.)  if -  =  -  =  -,  etc. 
a      b      c 

The  ratio  of  each  term  of  a  geometrical  progression  to 
the  preceding  term  is  called  the  common  ratio. 

The  following  are  examples  of  geometrical  progressions : 
1,  3,  9,  27,  etc., 

4,  2,  1,  J,    etc., 

and  I,         -1,  f,         -f,    etc. 

In  the  first  series  the  common  ratio  is  3,  in  the  second  series  it 
is  ^,  and  in  the  last  it  is  —  |. 

261.  If  the  first  term  of  a  G.P.  be  a,  and  the  cominon 
ratio  r,  then 

the  2d  term  will  be  ar, 
the  3d  term  will  be  ar^, 
the  4th  term  will  be  ar^, 

and  so  on,  the  index  of  r  being  always  less  by  unity  than 
the  number  giving  the  position  of  the  term  in  the  series. 
Thus  the  nth  term  is  ar"~^ 


GEOMETRICAL   PROGRESSION.  881 

We  can  therefore  write  down  any  term  of  a  G.P.  when 
the  first  term  and  the  common  ratio  are  given. 

For  example,  in  the  G.P.  whose  first  term  is  5,  and  whose 
common  ratio  is  3, 

the  4th  term  is  5  x  S^-i  =  5  x  33, 
and  the  10th  term  is  5  x  3io-i  =  5  x  3^. 

262.  When  any  two  terms  of  a  G.P.  are  given,  we  can 
find  the  first  term  and  the  common  ratio:  the  series  will 
thus  be  completely  determined. 

Suppose,  for  example,  that  the  5th  term  of  a  G.  P.  is  |  and 
the  7th  term  \\. 

Let  the  first  term  be  a,  and  the  common  ratio  r. 

Then  the  6th  and  the  7th  terms  are  ar^  and  ar^  respectively. 

Hence  ar*  =  | 

and  a»<  =  II ; 

.*.  by  division,  »^  =  I ; 

.*.  r  =  ±f. 
Then,  since  a  x  |f  =  f ,  a  =  |. 

Thus  the  series  is  f ,  ±  3,  2,  ±  f ,  etc. 

Ex.  The  4th  term  of  a  G.P.  is  189  and  the  6th  term  is  1701 ; 
find  the  8th  term. 

Let  the  first  term  be  a,  and  the  common  ratio  r  ]  then  the  4th 
term  will  be  ar*  and  the  6th  term  will  be  ai*. 
Hence,  by  the  conditions  of  the  problem, 

ar*  =  189  and  ar^  =  1701. 
By  division,  r^  =  9;  .-.  r  =  ±  3. 

Then  a  =  189/  (  ±  3)8  =  ±  7. 

Hence  the  8th  term  =  ar^  =  ±7(^±sy  =  15309. 

263.  When  three  quantities  are  in  geometrical  progres- 
sion, the  middle  one  is  called  the  geometric  mean  of  the 
other  two. 


382  GEOMETRICAL   PROGRESSION. 

Thus,  if  a,  b,  c  are  in  G.P.,  b  is  the  geometric  mean  of 
a  and  c.  By  the  definition  of  geometric  progression,  we 
have 

a      6' 
.•.  b^  =  ac; 

.-.  b  =  ±  Vac. 

Thus  the  geometric  mean  of  any  two  quantities  is  the 
square  root  of  their  product. 

It  should  be  noticed  that  quantities  which  are  in  con- 
tinued proportion  [Art.  246],  are  in  geometrical  pro- 
gression. 

264.  When  any  number  of  quantities  are  in  geometrical 
progression,  all  the  intermediate  terms  may  be  called 
geometric  means  of  the  two  extreme  terms. 

Between  two  given  quantities  any  number  of  geo- 
metric means  may  be  inserted. 

For  example,  to  insert  three  geometric  means  between  5  and  80. 

We  have  to  find  a  G.P.  with  three  terms  between  5  and  80,  so 
that  5  is  the  first  term  and  80  is  the  5th  term. 

Let  r  be  the  common  ratio  ;  then  the  5th  term  is  ar*. 
Hence  5r*  =  80;    .-.  r*  =  16;    .-.  r  =  ±2. 

Thus  the  series  is  5,   ±  10,  20,   ±  40,  80. 
The  required  means  are  therefore  ±  10,  20,   i  40. 

We  now  consider  the  most  general  case,  namely  to 
insert  n  geometric  means  between  a  and  b. 

We  have  to  find  a  G.P.  with  n  terms  between  a  and  6, 
so  that  a  is  the  first  and  b  is  the  {n  +  2)th  term  of  the 
G.P. 


is  a?-'*+^; 


GEOMETRICAL  PROGRESSION.  383 

Let  r  be  tlie  common  ratio.     Then  the  {n  -\-  2)th  term 

r       —  —  , 

Hence  the  required  means  are  ar,  ai^,  ar^,  •••,  ar",  where 


n+lj 

r 


EXAMPLES    LXXV. 

1.  Find  the  6th  terms  of  each  of  the  following  geometrical 
progressions : 

(i.)  9,  3,  1,  etc.       (ii.)  2,  -  3,  f,  etc.       (iii.)  a^,  ah,  b"^,  etc. 

2.  What  is  the  5th  term  of  the  G.P.  whose  first  term  is  3  and 
whose  3d  term  is  4  ? 

3.  The  3d  term  of  a  G.P.  is  1  and  the  6th  term  is  -  ^.     What 
is  the  10th  term  ? 

4.  The  4th  term  of  a  G.P.  is  .016  and  the  7th  term  is  .000128. 
What  is  the  first  term  ? 

6.   The  6th  term  of  a  G.P.  is  156  and  the  8th  term  is  7644. 
What  is  the  7  th  term  ? 

6.  The  3d  term  of  a  G.P.  is  2.25  and  the  7th  term  is  11.390625. 
What  is  the  5th  term  ? 

7.  The  3d  term  of  a  G.P.  is  4  and  the  6th  term  is  -  f    What 
is  the  10th  term  ? 

8.  The  4th  term  of  a  G.P.  is  y^  and  the  7th  term  is  —  j^-^. 
What  is  the  6th  term  ? 

9.  Find  the  geometric  mean  (i.)  of  4  and  9,  (ii.)  of  7  and  252, 
and  (iii.)  of  a^b  and  ab^. 


384  GEOMETRICAL    PROGRESSION. 

10.  Insert  two  geometric  means  between  1  and  —  8  ;  also  insert 
three  geometric  means  between  12  and  f ,  and  four  between  | 
and  f|. 

11.  Insert  five  geometric  means  between  3  and  .000192. 

12.  Insert  four  geometric  means  between  a^b~^  and  a~^¥. 

13.  The  sum  of  the  3d  and  4th  terms  of  a  G.P.  is  40,  and  the 
sum  of  the  6th  and  7th  teems  is  2560.     What  is  the  first  term  ? 

14.  The  sum  of  the  1st  and  2d  terms  of  a  G.P.  is  72,  and  the 
sum  of  the  3d  and  4th  terms  is  8.    What  is  the  first  term  ? 

15.  If  a,  b,  c,  d  are  in  G.P.,  show  that  ad  =  be. 

16.  If  all  the  terms  of  a  G.P.  be  multiplied  by  the  same  quan- 
tity, the  products  will  be  in  G.P. 

17.  Show  that  the  reciprocals  of  the  terms  of  a  geometrical  pro- 
gression are  in  G.P. 

18.  If  every  alternate  term  of  a  G.P.  be  taken  away,  the  remain- 
ing terms  will  be  in  G.P. 

19.  If  between  every  two  consecutive  terms  of  a  G.P.  their  geo- 
metric mean  be  inserted,  the  whole  will  form  another  geometrical 
progression. 

20.  Show  that  the  product  of  any  two  terms  of  a  geometrical 
progression,  which  are  respectively  equally  distant  from  the  first 
and  the  last  terms,  is  equal  to  the  product  of  the  first  and  last 
terms. 

265.  To  find  the  sum  of  any  number  of  terms  of  a 
geometrical  progression. 

Let  a  be  the  first  term,  and  r  the  common  ratio.  Let 
n  be  the  number  of  the  terms  whose  sum  is  required, 
and  let  I  be  the  last  term. 

Then,  since  I  is  the  nth-  term,  we  have 


GEOMETRICAL   PROGRESSION.  385 

Hence,  if  JS  be  the  required  sum, 

S=za-\-ar  +  ar^-\ \-  ar""-'^  +  ar'*-^ 

Multiply  by  r ;  then 

Sr  =  ar  +  ar^  -{- ar^  -\ h  ar""  ^  +  ar^. 

Hence,  by  subtraction, 

S  —  Sr  =  a— ar"" ', 
that  is  /S(l  -  r)  =  a(l  -  r»); 

1  —  r 

The  above  formula  may  be  written  in  another  form ; 
for,  by  substituting  the  value  I  =  ar**~^,  we  have 

e      a  —  rl 

Ex.  1.     Find  the  sum  of  10  terms  of  the  series  1,  2,  4,  etc. 
Here  a  =  1,  r  =  2,  n  =  10 ; 

o        1  —  r*» 


1-r 
1  -2^0 
1-2 


=  210  -  1  =  1023. 


Ex.  2.  Find  the  sum  of  six  terms  of  the  series 

2-3  +  I  -,  etc. 

—  3 
Here  a  =  2,  r=  — ,  and  n  =  6 ; 

\  —  r 

_gl-(-f)^-g !! 

~       l-(-f)    "      1  +  f 

2b 


386  GEOMETRICAL  PROGRESSION. 

266.  Sum  to  Infinity  of  a  G-eometrical  Progression.  It 
might  at  first  be  thought  that,  by  taking  a  very  great 
number  of  terms  of  any  geometrical  progression,  the 
sum  could  be  made  as  great  as  we  please ;  an  example 
will,  however,  show  that  this  is  not  the  case.  Suppose 
that  we  have  a  line  two  inches  long,  and  divide  it 
into  two  equal  parts  and  take  away  one;  then  bisect 
the  remainder  and  take  away  one  of  the  parts ;  and  sup- 
pose this  process  continued  to  any  extent.  Then  the 
lengths  in  inches  of  the  successive  parts  taken  will  be  1, 
1  1  \,  etc.  Now  it  is  clear  that  the  sum  of  all  the  parts 
taken  away  can  never  exceed  two  inches,  but,  as  the  part 
which  is  left  over  will  diminish  without  limit  as  the 
number  of  the  operations  is  increased,  the  sum  of  the 
parts  taken  away  can  be  made  to  differ  from  two  inches 
by  a  length  which,  is  smaller  than  any  conceivable  length; 
hence,  by  taking  a  sufficiently  large  number  of  terms, 

can  be  made  to  differ  from  2  by  an  arbitrarily  small 
quantity. 

From  Art.  265  we  have 

S  —  <^(^  — ^")  _     ct     _   gr" 
1  —  r         1  — r      1  —  r 

Kow  if  r  is  Si  proper  fraction,  whether  positive  or  nega- 
tive, the  absolute  value  of  r""  will  decrease  as  w  increases ; 
moreover  r"  can  be  made  as  small  as  we  please  by  suffi- 
ciently increasing  the  value  of  n. 

Hence,  when  r  is  numerically  less  than  unity,  the  sum 
of  a  very  large  number  of  terms  of  the  series  can  be 
made  to  differ  from  a/(l  —  r)  by  as  small  a  quantity  as 
we  please. 


GEOMETRICAL  PROGRESSION.  387 

It  is  customary  to  express  this  fact  in  conventional 
language  by  the  statement :  The  sum  to  infinity  of  the 
geometrical  progression  a  -^  ar  -{-  ar^  +  '•-,  in  which  r  is 
numerically  less  than  unity,  is  a/{l  —  r). 

Ex.  1.     Find  the  sum  to  infinity  of  the  series 

(i.)    1  +  ^  +  ^  +  ^  +  .... 

(ii.)  9-6  +  4...... 

In  (i.)  a  =  l,     r=J; 


•••^=^^.=^^i- 

In  (ii.) 

«=.  .=-«=-!; 

.:  S=     "     =         »         = 

9^27 
!-»•      l-(-f)      f      6* 

Ex.  2.    Find  the  value  of  .234. 
A.  .234  =  .23444. .. 

=  2.  +  A  +  A  +  A  +  A+etc. 
10      10'-2      103  ^  10*  ^106 

xr        4,4,        ,    .  «   .,         4/108       10^    4   _    4 

^^^IP  +  ioi+-'^^^^"^'y  =  W,=  9- ^10^-960* 
Hence  .234  =  ^^  +  ^k  +  ^h  =  Uh- 

Ex.  3.    Find  the  geometrical  progression  whose  sum  to  infinity 
is  18  and  whose  second  term  is  —8. 

Let  a  be  the  first  term  and  r  be  the  common  ratio. 

Then  the  second  term  is  ar,  and  the  sum  to  infinity  is  a/(l  —  r). 

Hence  ar  =  —  8, 

and  a/(l-r)=18. 


388  GEOMETRICAL   PROGRESSION. 

Hence,  by  division,  we  have 

r(l  _  r)  =  ^  ; 
^  ^      18  ' 

.♦.  r2-r:=f  ; 

.-.  r  =  -i,  or  r  =  f 

If  r  =  -l,    a  ==^  =  2i. 

3  r 

Thus  the  series  is  24,  —  8,  f ,  etc. 

The  value  r  =  |  is  inadmissible,  for  the  sum  of  a  geometrical 

progression  is  not  given  by  the  formula 

1  —r 
except  when  r  is  numerically  less  than  unity. 

EXAMPLES   LXXVI. 
Find  the  sum  of  the  following  series  : 

1.  8  +  12  +  18  +  27  +  •..  to  n  terms. 

2.  12  +  9  + 6f  +  •••  to6  terms. 

3.  a  +  -  +  -  +  —  to  n  terms. 

r      r^ 

4.  li  +  \\  +  l^V  +  •••  to  8  terms. 

5.  f  +  i  +  IH to  6  terms. 

6.  12-9  +  61 to  infinity. 

7.  4  +  1.2  +  .36  +  .108  +  ...  to  infinity. 

8.  4  + .8  + .16+ •••  to  infinity. 

9.  a2_^  055  _^  52  _^  ...  to  w  terms. 

10.  3  -  2  +  I  -  •••  to 'infinity. 

11.  Find  the  sum  of  the  successive  powers  of  2  from  2  up  to 
4096  inclusive. 

12.  Find,  correct  to  four  places  of  decimals,  the  sum  to  infinity 

of  the  series  1  +  —  +  -  +  ••.. 
■^2      2 


GEOMETRICAL  PROGRESSION.  389 

13.  Find  the  geometric  mean  of  4  x^  — 12  x4-  9  and  9  x^ + 12  x + 4. 

14.  Show  that  the  product  of  any  odd  number  of  terms  of  a 
G.P.  is  equal  to  the  nth  power  of  the  middle  term,  n  being  the 
number  of  the  terms. 

16.  Find  the  G.P.  whose  sum  to  infinity  is  4,  and  whose  2d 
term  is  f . 

16.  Find  the  G.P.  whose  sum  to  infinity  is  9,  and  whose  2d 
term  is  —  4. 

17.  The  sum  of  the  first  10  terms  of  a  certain  G.P.  is  equal  to 
33  times  the  sum  of  the  first  5  terms.     What  is  the  common  ratio  ? 

18.  If  the  sum  of  a  geometrical  series  to  infinity  is  n  times  the 

first  term,  show  that  the  common  ratio  is  1 

n 

19.  If  the  common  ratio  of  successive  terms  of  a  G.P.  be  posi- 
tive and  less  than  ^,  show  that  each  term  is  greater  than  the  sum  of 
all  that  follow  it. 

20.  The  6th  teim  of  a  G.P.  is  8  times  the  3d  term,  and  the  sum 
of  the  first  two  terms  is  24.     Find  the  series. 

21.  Find  the  sum  to  infinity  of  the  series 

_«_+f_«_y+f_«_y+... 

a  +  b     \a-hbj       \a  +  bj 

22.  Find  the  sum  to  infinity  of  the  series 

23.  Show  that,  if 

Si  =  a  +  ar  +  ar^  +  ar*  +  —  +  ar^ 

and  S2  =  a  —  ar-\-  ar^  —  ar^  + h  ar2»» ; 

then  S1S2  =  a"^  +  a^ -{■  ah^  +  —  +  a^r^. 

24.  Sum  the  series  VI  +  iV^  +  f  Vl  +  ••*  ^o  infinity. 

25.  The  6th  term  of  a  G.P.  is  jslj^  and  the  2d  term  is  ^j. 
Find  the  sum  to  infinity. 


390  GEOMETRICAL  PROGRESSION. 

26-  Find  three  numbers  in  G.P.  such  that  their  sum  is  14,  and 
the  sum  of  their  squares  is  84. 

27.  The  sum  of  the  1st,  2d,  and  3d  terms  of  a  G.P.  is  42,  and 
the  4th  term  exceeds  the  first  by  126.     What  is  the  series  ? 

28.  The  sum  of  the  1st,  2d,  and  3d,  terms  of  a  G.P.  is  to  the 
sum  of  the  3d,  4th,  and  5th  terms  as  1:4,  and  the  5th  term  is  8. 
What  is  the  series  ? 

29.  Show  that,  if  a,  6,  c,  d  be  in  G.P.,  then  will  a  +  ?),  6  +  c, 
c  +  d,  and  also  a^  +  h\  h"^  +  c2,  c^  +  d^  be  in  G.P. 

30.  Show  that,  if  a,  6,  c  be  the  jDth,  gth,  and  rth  terms  respec- 
tively of  a  G.P.,  then  will  a?-'-6'-i>cP-2  =  1. 

31.  Show  that,  if  (ofi  +  62)  (52  +  ^2)  =  {ah  +  6c) 2,  then  will  a,  6, 
c  be  in  G.P. 


267.  Sometimes  we  are  not  told  the  law  which  con- 
nects successive  terms  of  a  series;  but  when  a  certain 
number  of  the  terms  are  given,  the  law  can  in  simple 
cases  be  at  once  determined. 

Suppose,  for  example,  we  have  the  series 
3,    9,     15,  etc. 

Here  9-3  =  6,  and  15  -  9  =  6. 

Thus,  the  series  is  an  arithmetical  progression^  whose  common 
difference  is  6. 

Again,  in  the  series  3,  9,  27,  etc., 

Since  9  —  3  =  6,  and  27  —  9  =  18,  the  series  is  not  an  arith- 
metical progression.  We  then  see  whether  it  satisfies  the  condi- 
tions for  being  a  geometrical  progression,  namely  |  =  ^-^^  which  is 
the  case.  Thus  the  series  is  a  geometrical  progression^  whose 
common  ratio  is  3. 


GEOMETRICAL  PROGRESSION.  391 

EXAMPLES  LXXVII. 
Sum  the  following  series  : 

1.  3,  2.7,  2.4,  etc.,  to  21  terms. 

2.  2,  18,  162,  etc.,  to  7  terms. 

3.  1,  .2,  .04,  etc.,  to  infinity. 

4.  a,  6,  — ,  etc.,  to  n  terms. 

a 

6.   .3,  .03,  .003,  etc.,  to  infinity. 

6.  3  +  4.3  +  5.6  +  ..•  to  11  terms. 

7.  a  +  i«_±^  +  5^±2^+...tol9terms. 

3  3 

g.   J  _  ^  +  ^  _  etc.,  to  8  terms. 
9.  3J  +  1^  +  ^  +  etc.,  to  infinity. 

10.  2J  +  ej  +  15|  +  •••  to  5  terms. 

11.  Sum  the  following  series  to  6  terms,  and  when  possible  to 
infinity : 

(i.)  2^  +  6^  +  10  +  ...  (iii.)4-34-f--  (v.)i  +  i  +  f  +  - 
(ii.)  9  +  6  +  1  +  ...         (iv.)  I  +  i  +  I  +  ...  (vi.)  ^  -  1  +  I  _  ... 

12.  Show  that,  if  an  odd  number  of  quantities  are  in  geomet- 
rical progression,  the  first,  the  middle,  and  the  last  of  them  are 
also  in  geometrical  progression. 

13.  The  sum  of  the  first  7  terms  of  an  A.P.  is  49,  and  the  sum 
of  the  next  8  terms  is  176.     What  is  the  series  ? 

14.  The  arithmetic  mean  of  the  1st  and  3d  terms  of  a  geo- 
metrical progression  is  5  times  the  2d  term.  Find  the  common 
ratio. 

16.  If  the  4th  term  in  an  arithmetical  progression  is  the 
geometric  mean  of  the  2d  and  7th  terms,  show  that  the  6th  term 
is  the  geometric  mean  of  the  2d  and  14th. 


392  GEOMETRICAL  PROGRESSION. 

16.  If  P  be  the  continued  product  of  n  terms  of  a  geometrical 
progression  whose  first  term  is  a  and  last  term  I,  show  that 

P2=(a?)n. 

17.  The  continued  product  of  three  numbers  in  G.P.  is  216,  and 
the  sum  of  the  products  of  them  in  pairs  is  156.    Find  the  numbers. 

18.  Divide  25  into  five  parts  which  are  in  A.P.,  and  which  are 
such  that  the  sum  of  the  squares  of  the  least  and  greatest  of  them 
is  one  less  than  the  sum  of  the  squares  of  the  other  three. 

19.  Insert  between  6  and  16  two  numbers  such  that  the  first 
three  may  be  in  A.P.  and  the  last  three  in  G.P. 

20.  If  a,  b,  c  be  in  geometrical  progression,  and  x,  y  be  the 
arithmetic  means  between  a,  6,  and  6,  c,  respectively,  prove  that 

?  =  !  +  !,   and2  =  «  +  -». 
h     z     y  X     y 


HARMONICAL  PROGRESSION.  393 


CHAPTER   XXVI. 
Harmonical  Progression.  Other  Simple  Series. 

268.  A  series  of  quantities  is  said  to  be  in  Hannonical 
Progression  when  the  difference  between  the  first  and  the 
second  of  any  three  consecutive  terms  is  to  the  difference 
between  the  second. and  the  third  as  the  first  is  to  the 
third. 

Thus  a,  h,  c,  dy  etc.,  are  in  harmonical  progression,  if 

a  —  6:6  —  c  ::  a'.  Cy 

b  —  c:  c  —  d: :  b  :  dj 
and  so  on. 


269.   If  a,  6,  c  are  in 

harmonical  progression, 

we 

have 

by  definition 

a  — 

6:6 

—  c::a:c. 

Hence 

c{a- 

-6): 

=  a(6-c), 

or 

ca- 

-be. 

=  ab  —  ac. 

Divide  by  abc ; 

then 

1 

_1 

1     1 

6 

a 

"c-6' 

which  shows  that 

1 

a' 

1    1 

6'  c 

are  in  arithmetical  progression. 

Thus,  if  quantities  are  in  harmonical  progression,  their 
reciprocals  are  in  arithmetical  progression. 


394  HARMONICAL  PROGRESSION. 

The  above  relation  between  quantities  in  harmonical 
progression  is  much  more  frequently  employed  than  that 
given  in  the  preceding  article. 

270.  When  three  quantities  are  in  harmonical  pro- 
gression, the  middle  one  is  called  the  harmonic  mean  of 
the  other  two. 

If  a,  b,  c  are  in  harmonical  progression, 
111 
a   b   c 
are  in  arithmetical  progression ; 

"  b      a     c      &' 

b     a     c 


a-\-  c 

Thus  the  harmonic  mean  of  any  two  quantities  is  twice 
their  product  divided  by  their  sum. 

271.  If  we  put  A,  G,  H  for  the  arithmetic,  the  geo- 
metric, and  the  harmonic  means  respectively  of  any  two 
quantities  a  and  b,  we  have 

A^Ua-\-b),   G  =  Vab,  and  ^=-^-^^ 

Hence,  AxH=\{a  +  b)  x-?-^  =  a&; 

a  -\-  0 

.'.  AH=  0\ 
Hence  O  is  the  geometric  mean  of  A  and  H. 
Thus  the  geometi^ic  mean  of  any  two  quantities  is  also 
the  geometric  mean  of  their  arithmetic  and  harmonic  means. 


HARMONICAL  PROGRESSION.  395 

272.  Many  questions  concerning  quantities  in  har- 
monical  progression  can  be  solved  by  considering  the 
arithmetical  progression  whose  terms  are  the  reciprocals 
of  the  terms  of  the  harmonical  progression. 

For  example,  to   insert  n  harmonic   means   between 

any  two  quantities  a  and  b. 

ft  1  -I 
Insert  n  arithmetic  means  between  -  and  —     These, 

1  b 

by  Art.  258,  will  be 

l_^l/b-l/a^l_^,l/b-l/a^^^^ 
a  n-\-l        a  n  +  1 

Thus         1,   l  +  lA^»,l+2l/^^,...,l 
a    a         w+1        a  n+l  6 

are  in  A.  P. 

That  is 

1      (nb  +  a)      (n-l)b  +  2a         1 

a    {n-\-l)ab'       {n  +  l)ab     '  '"' b 

are  in  A.P. 

Hence  the  reciprocals  of  these  are  in  H.P. 

Thus  the  required  harmonic  means  are 

(n-\-l)ab        {n-{-l)ab        ^^^ 
nb-^a    '   (n-l)6  +  2a' 

Note.  —  It  is  of  importance  to  notice  that  no  formula  can  be 
found  giving  the  sum  of  any  number  of  terms  in  harmonical 
progression. 

273.  The  conditions  that  a,  b,  c  may  be  in  A.P.,  G.P., 
or  H.P.,  respectively,  may  be  written 

a  —  b:b  —  c:.a:a, 
a  —  b:b  —  c::a:by 
a  —  b:b  —  c::a:c. 


396  HARMONICAL   PROGRESSION. 

The  first  and  third  follow  from  the  definitions  of  A.R 
and  H.P.,  and  the  second  can  be  easily  verified  by  mul- 
tiplication. 

Ex.  1.  The  second  term  of  a  harmonical  progression  is  2,  and 
the  fourth  term  is  6  ;  find  the  series. 

The  second  term  of  the  corresponding  arithmetical  progression 
is  ^,  and  the  fourth  term  is  ^.  Hence,  if  a  be  the  first  term,,  and 
d  the  common  difference,  we  have 

a-\-  d  —  \^  and  a  +  Zd  =  \, 
whence  «  =  I,  and  d=  —  ^. 

Thus,  the  arithmetical  progression  is 

f>     5>     3'     o'  ®tc. 
Hence  the  harmojiical  progression  is 

f,    2,    3,    6,  etc. 

Ex.  2.  Show  that  a  +  6,  2  6,  and  6  +  c  will  be  in  H.P.  if  a,  6,  c 
are  in  G.P. 

The  three  quantities  a  +  6,  2  5,  and  6  +  c  will  be  in  H.P.  if 
1.1  2 


a  +  6      6  +  c     2h 

or  if 

b(b  +  c)+  b(a  +  &)  =  («  +  b)(b  +  c)  ; 

that  is,  if 

62  4-  6c  +  a6  +  62  =  a6  +  62  +  ac  4-  6c, 

or  if 

62  =  ac ; 

and  this  is  the  case  when  a,  6,  c  are  in  G.P. 

EXAMPLES  LXXVIII. 

1.  Show  that  if  the  terms  of  a  harmonic  progression  be  all 
multiplied  by  the  same  quantity,  the  products  will  be  in  harmonical 
progression. 

2.  Insert  5  harmonic  means  between  1  and  7. 

3.  Insert  4  harmonic  means  between  f  and  f . 

4.  Show  that,  if  the  arithmetic  mean  of  two  quantities  is  1, 
their  harmonic  mean  will  be  the  square  of  their  geometric  mean. 


HARMONICAL   PROGRESSION.  397 

5.  Show  that,  if  a^,  52^  ^^  are  in  arithmetic  progression,  6  +  c, 
c  +  o,  and  a  +  b  will  be  in  harmonical  progression. 

6.  Show  that,  if  x,  y,  ^  be  in  H.P., 

(2/  +  0-x)2,    (0  +  x-2/)2,    (x  +  y-zy 
will  be  in  A.P. 

7.  Show  that,  if  x,  y,  0  be  in  H.P.,  then  will  — ^,  — ^» 

—=- — ,  be  also  in  H.P. 

a^  +  y 

8.  Show  that,  if  x,  y,  z  be  in  H.P.,  then  will 

X  y  z 

y  +  z  —  x'  z  +  x  —  y'  x-\-y  —  z 

be  also  in  H.P. 

9.  If  a,  6,  c,  d  be  in  H.P.,  then  will ,  ^ , 

c  ^         <i         V    •    TJX.  6  +  c  +  d   c  +  d+a 

,  and be  in  H.P. 

d-h  a-\-  b  a-\-b  -\-  c 

10.  If  a,  b,  c  be  in  H.P.,  then  will  2a  —  b,  b,  and  2c  —  6  be 
in  G.P. 

11.  If  a,  6,  c  be  in  H.P.,  then  will  a,  a  —  c,  and  a  —  6  be  in  H.P. 

12.  Show  that 

(rt2  +  62)  (<j2  +  a6  +  62)^  ^4  4.  ^252  +  54^  ^ 

and  (a2  +  b'^)  {a^  -  ab -\-  62) 

are  in  H.P. 

13.  Show  that,  if  x,  ai,  02,  y  be  in  A.P.,  x,  gi,  92,  V  in  GJP., 
and  X,  ^1,  ^,  y  in  H.P.,  then  will 

giQi  _  ai  +  q2. 

14.  If  a,  6,  c  be  in  H.P.,  then  will 

b+  a _^  6  +  c _ 2_ 
6  —  a     5  —  c 

15.  If  a,  6,  c  be  in  G.P.,  then  will  -^— ,  -,  and  — ?—  be  in  A.P. 

a  +  6   6  6  +  c 


398  HARMONICAL   PROGRESSION. 

16.  If  a,  b,  c  be  in  A. P.,  and  6,  c,  d  in  H.P.,  then  will 

a:b  =  c:d. 

17.  Show  that,  if 

a     c     h  —  a     h  —  c 
then  6  =  a  +  c,  or  else  a,  &,  c  are  in  harmonical  progression. 

18.  If  «,  6,  c,  d  be  in  H.P.,  then  will 

3(6  -  a){d  -c)  =  (c-  b){d  -  a). 

19.  Show  that,  if  a,  6,  c  be  in  H.P.,  then 
2         1^1 


b     b  —  a     b  —  c 

20.  Three  numbers  are  in  A.P.  The  product  of  the  extremes 
is  five  times  the  mean,  and  the  sum  of  the  two  largest  is  three  times 
the  least.    Pind  the  numbers. 

21.  Show  that,  if  ±±A^   b,   ^-±-^  are  in  A.P.,  then  a,  -,   c 

1  —  ab         1  —  be  b 

will  be  in  H.P. 

22.  If  a,  6,  c  be  in  A.P.,  and  a^,  b"^,  c^  in  H.P,,  prove  that 
—  -,  b,  c  are  in  G.P.,  ov  a  =  b  =  c. 

23.  Three  numbers  are  in  A.P,,  and  if  1  be  added  to  the  greatest, 
they  will  be  in  G.P. ;  show  that  the  smallest  is  equal  to  the  square 
"iiZ  the  common  difference. 

24.  Show  that,  if  x,  a,  6,  y  be  in  A.P.,  and  x,  w,  v^  y  in  H.P., 
then  av  =  bu  =  xy. 

25.  Show  that,  if  ai,  ^2,  as,  04  be  in  H.P.,  then 

aia2  +  0,2(23  +  cisdA  =  3  aiGi. 

26.  Show  that,  if  a,  6,  c  be  in  A. P.,  and  a,  c,  b  be  in  G.P.,  then 
will  &,  a,  c  be  in  H.P. 

274.  Other  Simple  Series.  There  are,  besides  the  pro- 
gressions, many  other  series  the  successive  terms  of  which 
are  formed  according  to  simple  laws.  The  following  are 
examples  of  such  series : 


SIMPLE  SERIES.  399 

12  4.  22  + 32 +...+^2, 

1 .  2 -f  2  .  3  H- 3  .  4  +  ...  +  w(ri  + 1), 
and  J_  +  J_-f  J_+...+        1 


1.22.33.4'  w(n  +  l) 

275.  We  proceed  to  find  the  sum  of  n  terms  of  some 
of  these  simple  series. 

The  sum  of  n  terms  of  a  series  is  generally  denoted 
by  Sn,  and  the  sum  of  the  series  when  continued  without 
limit  by  S^. 

Ex.  1.   Find  the  sum  of  n  terms  of  the  series 
1. 2  +  2. 3  +  3. 4  +  4.5  +  . .. 
Here 

>S„  =1 . 2  +  2 . 3  +  3 . 4  +  ...  +  (n  -  l)n  +  n(n  +  1). 
Let 
S'=  1 .  2  .  3  + 

2.3-4  +  3-4-5  +  ..-  +(n-l)n(n  +  1)+ n(M  +  l)(n +2). 
Shifting  each  term  one  place  to  the  right,  we  have 
2  =  1.2-3 +  2.3-4  +  ...  +(n-2)(n-l)n+(n-l)n(n  +  l) 

+  n(n+-l)(7i  +  2). 

Now  subtract  the  last  result  from  the  preceding,  taking  each 
term  from  the  one  above  it ;  then  we  have 
0  =  1  -  2  -  3  +  3  -  2 .  3  +  3 . 3  -  4  +  ...  +  3(n  -  l)n  +  3  n(n  +  1) 

-n(w  +  l)(n +  2). 

Hence 

3{1  - 2  +  2  -  3  +  3  -  4  +  -  +  (w  -  l)ri  +  n(n+l)}=  n(n+l)(w  +  2)  ; 
.-.  Sn  =  inin-\-l)in  +  2). 

Ex.  2.  Find  the  sum  of  n  terms  of  the  series 

1-2.3  +  2-3-4  +  3.4-5  +  .... 
Here 

/§;  =  1.2-3+2. 3-4+3-4- 5  +  .-  +  (n-l)n(n+l)  +  n(n+l)(n+2). 


400  SIMPLE  SERIES. 

Let 
S  =  1.2.  3.  4 +  2.  3.  4- 5 +  3.  4.5.6 +  ••• 

+  (w  -  l)n(n  +  l)(n  +  2)+  n(n  +  l)(w  +  2)(w  +  3). 
Then 

S=  1. 2.  3. 4  +  2. 3.4.5+. •• 

+  (w  -  2)(w  -  l)w(n  +  1)  +    (n-  l)w(w  +  l)(n  +  2) 
+  n(w  +  l)   (n  +  2)(/i  +  3). 
Hence,  subtracting  each  term  from  the  one  above  it, 
0  =  1.2.3.4  +  4.2.3.4  +  4.3.4.5+... 

+  4(w-l)w(w  +  l)+4n(w+  l)(w  +  2) 
-w(?i  +  l)(w  +  2)(n  +  3); 
.-.  4(1.2.3  +  2.3.4  +  3.4.5+...  +  n(w+  l)(w  + 2)} 

=  n(w+  l)(7i  +  2)(n  +  3); 
.-.  ^„  =  jn(w  +  l)(n  +  2)(n  +  3). 

Note.  —  This  series  and  the  preceding  are  examples  of  an  impor- 
tant type  of  series  in  which  (i.)  each  term  contains  the  same 
number  of  factors,  (ii.)  the  factors  of  any  term  are  in  A.P.,  and 
(iii.)  the  first  factors  of  the  successive  terms  form  the  same  A.P. 
as  the  successive  factors  of  the  first  term. 

The  S  series  by  means  of  which  the  required  sum  can  always  be 
found  is  the  series  whose  terms  are  formed  according  to  the  same 
law  but  with  an  additional  factor  at  the  end. 

Ex,  3.    To  find  the  sum  of  n  terms  of  the  series 

12+22  +  32  +  ..-. 
Here  /^„  =  P  +  22  +  32  +  ...  +  w^. 

Now  n^  =  n{n  +  1)  —  n  ; 

.-.  /S'„  =  1.2  +  2.3  +  ... +  w(w+l)-l-2 n. 

But,  by  example  1, 

1.2  +  2.3  +  3-4+  ...  +  n(w  +  l)=^w(w +  l)(n  + 2); 
also  1 +2 +  3+ ...  +  w  =  ^w(w  + 1). 

Hence  8n  =  \n{n-\- V){n +  2)-  ^n{n  +  \) 

=  in(w+  l)(2w+  1). 


SIMPLE   SERIES.  401 

Ex.  4.   To  find  the  sum  of  n  terms  of  the  series 

13  +  23  +  33+... +  n8. 

Here  Sn  =  l^ +  2^  +  Z^ -h  —  +  n\ 

Now  4  w3  =  {w(w  +  1)}2  —  {(n  —  l)n}2,  for  all  values  of  n. 

Hence  we  have 

4. 13  =(1.2)2, 

4 .  28  =  (2  .  3)2  _  (1 . 2)2, 

4.33  =(3.4)2 -(2.3)2, 


4(n  -  1)8  ==  {(w  -V)nf-  {(n  -  2)  (w  -  \)f, 
4  w3  =  {n(n  +  1)}2  -  {(n  -  \)nf. 
Hence,  by  addition, 

4(18  +  23  +  38  +  ...  +  w8}  =  {n{n  +  \)f, 

.-.         13+28  +  38+...  +  w3   =  I  n2(w+  1)2. 

This  result  may  be  expressed  differently ;  for  since 

l  +  2  +  3  +  ...  +  n  =  Jw(w  +  1), 

we  have       18  +  28  +  38  +  ...  +  n8  =  (1  +  2  +  3  +  ...  +  n)2. 

Thus  the  sum  of  the  cubes  of  the  first  n  integers  is  equal  to  the 
square  of  their  sum. 

Ex.  6.  Find  the  sum  of  n  terms  of  the  series 

-JL  +  ^  +  _JL+.... 
1.2    2.3    3.4 

Here   ;S^„  =  J_ +  -1- +  J-+ ...  + 


1-2     2.3     3.4  («-l)n     n(n  +  l) 

Let  2  =  1  +  1  +  1  +  1  +  ...  +  !  + 


1234  n     n+1 

then  S=        1  +  1  +  1  +  .. .  +  _1_  +  1  + 


123  n-lww+1 

2o 


402  SIMPLE   SERIES. 


Hence,  by  subtraction, 

0  =  1-    111  -1  11 


1-2      2-3     3-4              (n-l)»i      w(n  + 1)      n  +  1 
Hence        J_ +_!_....+ 1 =  i_     1 


1-2      2-3  w(n  +  l)  w  +  1 

When  n  is  infinite,  is  zero ;  hence  the  sum  to  infinity  of 

n  +  1 

the  series 1 1 +  •■•  is  1. 

1-2     2.3     3-4 

Ex.  6.    Find  the  sum  of  n  terms  of  the  series 

1      +__!      +      1^+.... 


1.2.3      2.3.4     3.4.5 

Take  a  series  S  formed  according  to  the  same  law  but  with  one 
factor  less  in  each  term,  and  proceed  as  before.     The  result  is 


2  11 -2      (n  +  l)(n  +  2)/ 


2 

Ex.  7.   Find  the  sum  of  n  terms  of  the  series 
l-\-2x  +  Sx^  +  4x^+  .... 
Let  /S'  =  1  +  2  a;  +  3 x2  +  4 ic3  +  ...  +  nx^-'^  +  (w  +  l)a?». 
Then        Sx  =  x  +  2  x'^  +  S  x^  +  "•  ■{•  wx"  +  (n  +  l)a?»+i. 
Hence,  by  subtraction, 

S(l-x)=l  +  x  +  x'^-h  x^ -{-'•'  +x^-(n  +  l)a;"+i. 

But  1 +  a;  + a;2+ ...  4-cc«  =  ^!^ 

1  —X 

Hence,  S(l  -x)=  Lz^!!!^  _  (n  +  l)a;«+i ; 


^  =  1^^;_(,  +  1)^+^ 


ii-xy    "       'i-x 


MISCELLANEOUS   EXAMPLES   VI.  403 

EXAMPLES  LXXIX. 

Find  the  sum  of  the  foUowmg  series  to  n  terms,  and  when 
possible  to  infinity. 

1.  2.4  +  4.6  +  6.8  +  ....  3.    1.4  +  4.7  +  7.10  +  .... 

2.  3.5  +  5.7  +  7.9  +  ....  4.   2. 6  +  6. 8  +  8- U  +  -. 

5.  1.3.5  +  3.5.7  +  5.7.9  +  .... 

6.  2  .  7  .  12  +7  .  12  .  17  +  12  .  17  .  22  +  .... 

7.  -i-  + -1-+  — +  ...  8.    — +  -i-+-^  +  .... 
1.33.55.7  2. 66. 8     8. 11 

9.  _JL_  +  _l_  +  _i_+.... 

1.3.53.5.76.7.9 

10.  _L_  +  _^-  +  __L_  +  .... 

2.5.8      6.8.11     8-11.14 

11.   \ + 1 + I +.... 

(a;  +  l)(x  +  2)      (x  +  2)(a:  +  3)      (x  +  3)(a;  +  4)^ 

12.   ^ + ^ ^ 1 +  .... 

(l+x)(l+2a;)      (l+2a;)(l+3a;)      (l+3a;)(l+4a;)  ^ 

13    1  I      1      I  1  I  ^  I 

*    11+2      1+2+3      1+2+3+4 

14.    12  +  32  +  52  +  ....  15.    18  +  38  +  58  +  .... 

16.   a(a  +  &)  +  (a  +  6)(a  +  2  6)  +  (a  +  2  6)(a  +  3r6)+.... 


MISCELLANEOUS   EXAMPLES  VL 

A.   1.    Simplify 
(y+3)(2^-l)-3(y+l)(y2-9)+3(y-l)(y2_9)-(2,-3)(y2_i). 

2.  Prove  that 
(2y-x)»-(2x-y)»      (2y-x)H(2x-y)»  ^  ^^   ,_ 

3(y-a;)  x+y  "        " 


404  MISCELLANEOUS   EXAMPLES   VI. 

3.  Find  the  H.C.F.  of  Sx^  -  ISx^  +  2Sx  -  21  and  6x^-\-x^ 
-44x  +  21. 

For  what  value  of  x  will  both  expressions  vanish  ? 

4.  Solve  the  equations  : 

(i.)   Sx  +  --l  =  12x  +  ^+U  =  ^-2x-U. 

y  y  y 

(ii.)   x2  -  (a  +  6  +  2  c)a:  +  («  +  6  +  c)c  =  0. 

5.  A  man  buys  9  oxen  and  20  sheep  for  £230 ;  by  selling  the 
oxen  at  a  gain  of  25  per  cent,  and  the  sheep  at  a  loss  of  20  per 
cent,  he  gains  £  35  altogether.    Find  the  price  he  gave  for  each. 

6.  If  a  and  yS  be  the  roots  of  the  equation  a;^  +  4  x  +  3  =  0, 

show  that  the  equation  whose  roots  are  "  "^      and  °  "^  ^   is 

a  iS 

3  a:2  -  16  a;  +  16  =  0. 

7.  Multiply  ah-^-2  ai6-^  +  4-8  a-^&^+16a-H  by  a^+2h^. 

8.  Rationalize  the  denominator  of  -^ — ^, 

5-4V3 

and  simplify  V25  +  4^34. 

9.  Sum  the  series  5  —  3|  +  2i|—  etc.,  to  8  terms,  and  find  its 
limit  when  continued  to  infinity. 

10.  Insert  20  arithmetic  means  between  100  and  300,  and  find 
their  sum. 


B.    1.    Simplify  Z{x  -  2{y  -  z)]-l'iy  +  2{x  -  y  -  z}']. 

2.  Find  the  factors  of 

wnx2  +  m'^xy  +  n'^xy  +  mny^^ 
and  of  x^  —  x'^y  +  xy^  —  y^. 

3.  If  the  sum  of  two  numbers  be  equal  to  4,  show  that  their 
difference  is  one-quarter  of  the  difference  of  their  squares. 

4.  Solve 

(i.)  -^-+-^  =  1. 
x  +  a     x  +  b 


(ii.)  V12  X  -  3  +  Va:  +  2  +  V7  X  -  13  =  0. 


MISCELLANEOUS  EXAMPLES   VI.  405 

6.   Show  that  x^  —  8  x  +  22  can  never  be  less  than  6. 

6.  Find  the  square  root  of  a^  +  2  a^h"^  -  2  a^c  +  ft*  -  2  hH  +  c\ 

7.  If  a :  6  : :  c :  (?,  show  that 

Ct2  +  62  .  c2  +  (J2  .  .(oj  +  5)2  .(c  +  d)2. 

8.  Sum  the  following  series : 

(i.)  5  -  1  -  7  -  etc.  to  20  terms. 
(ii.)  2J  +  1  +  f  +  •••  to  infinity. 

9.  Find  the  sum  of  all  the  numbers  which  are  less  than  1000, 
and  are  divisible  by  7. 

10.  A  train  travelling  at  the  rate  of  37  J  miles  an  hour  passes  a 
person  walking  on  a  road  parallel  to  the  railway  in  6  seconds; 
it  also  meets  another  person  walking  at  the  same  rate  as  the  other, 
but  in  the  opposite  direction,  and  passes  him  in  4  seconds.  Find 
the  length  of  the  train. 

C.    1.  Find  the  value  of  V(y^  +  ^)  _  ^  -  <y  -  ^), 
V(«  -\-y)      x-z(y-  X) 

when  x  =  0,    y  =  1,    z  =  li. 

2.   Divide  4x*  -  9xV  -|-  I2xy^  -  4y*  by  2x2  +  3xy  -2y'^. 

8.   Show  that  the  square  of  the  sum  of  two  consecutive  numbers 
is  equal  to  four  times  their  product  increased  by  unity. 

4.   Find  the  L.  C.  M.  of  x^  -  6  ax  +  9  a^,   x^-ax-6a^  and 
3x2-12a2. 

6.    Simplify        _2_  +  ,^_+     ^  ^ 


a-2      1 -  a     a+ 1      a  +  2 

6.  If  40  minutes  would  be  saved  in  a  journey  by  increasing  the 
rate  of  the  train  by  5  miles  an  hour,  and  1  hour  would  be  lost  by 
diminishing  it  by  the  same  amount,  find  the  rate  of  the  train  and 
the  length  of  the  journey. 

7.  Simplify 

2  +  ^8  +  V2  -  V27  -  Vl2  +  V76  -  V(19  +  ^  V2)- 


406  MISCELLANEOUS   EXAMPLES   VI. 

8.  Prove  that  any  ratio  is  made  nearer  to  unity  by  adding  the 

same  number  to  each  of  its  terms.    Also  show  that  ^^  "^  ^^  is 

.  mh  -\-nd 

intermediate  between  -  and  -• 
h  d 

9.  Find  the  sum  of  the  following  series  : 

(i.)  21  +  15  +  9  +  ...  to  8  terms, 
(ii.)     5  +    2  +  .8  +  ...  to  infinity. 

10.   The  arithmetic  mean  of  two  numbers  is  17,  and  their  geo- 
metric mean  is  15.     What  are  the  numbers  ? 

D.    1.    Simplify  {x{x  +  a)-  a(x  -  a)}{x(x  -a)-  a(a  -  x)]. 

2.  Showthat  (w  + 1)*- w4=(2n  + l)(2w2  +  2w+ 1), 
and  that  (a  +  6)*  -  a*  -  6*  =  4  ab(a  +  6)2  -  2  a'^b'^. 

3.  Simplify     |i_3a:-20|r^_8a:-42|, 

4.  Solve  the  equations : 

(i.)  §-^  =  21,  ^-f  Z  =  -ll. 
X     y  X     y 

(ii.)  3x2  -  4a;?/  +  2  ?/2  =  33,  x"^  -  y"^  =  16. 

5.  A  number  consists  of  two  digits,  of  which  that  in  the  unit's 
place  is  the  greater;  the  difference  between  the  squares  of  the 
digits  is  equal  to  the  number,  and  if  the  digits  were  inverted  the 
number  thus  formed  would  be  7  times  the  sum  of  the  digits.  Find 
the  number. 

6.  Find  the  equation  whose  roots  are  the  cubes  of  the  roots  of 
the  equation  a;2  —  4  x  +  2  =  0. 

7.  Find  the  square  root  of  x^  -  2  x2  +  8  +  x-2  -  8  x"*  +  16  x'^. 

8.  If  a  :  &  :  :  c  :  c?,  prove  that  ^^^-±^  =  (^Lzi^V. 

c^-\-d^     \c-dj 

9.  Sum  the  following  series : 


(ii.)  9-6  +  4 to  infinity. 


MISCELLANEOUS   EXAMPLES   VI.  407 

10.  The  series  of  odd  numbers  is  divided  into  groups  as  follows : 
1 ;  3,  5  ;  7,  9,  11 ;  13,  15,  17,  19  ;  and  so  on.  Show  that  the  sum 
of  the  numbers  in  the  nth  group  is  n^. 

E.   1.  Simipmyx:2-2xy+y^-ix^-hxy-{-y^)-{x(-2y+x)+y^}. 

2.  Divide6a4+4&*-a5&  +  IS  ab^ +2 a-b"^  by  2  a^  +  4^ b"^  -  Sab. 

3.  Find  the  highest  common  divisor  and  the  lowest  common 
multiple  of  3  a*  -  3  a^b  +  ab'^  -  b^  and  4  a^  -  6  a6  +  6^, 

4      ^  x-1 


4.  Simplify  - 


1         1 
x—1      X 

6.   If  a  and  /3  are  roots  of  x^ -\-  mx -\-  n  =  0,  find  a^/S  +  /S^a  in 
terms  of  m  and  n. 

Test  your  result  on  the  equation  ic^  —  3  a;  +  2  =  0. 

6.  Find  the  value  of  x  for  which  Sx^  -{-  5  x  -\-  S  has  its  least  pos- 
sible value,  and  show  that  the  least  value  is  ||. 

7.  Show  that  if  a,  fe,  c,  d  be  in  continued  proportion, 

(a-  by^a 
[b  -  c)      d 

%.  Find  the  square  root  of 

4  x^  -  12  xy^  -7x^y  +  24  x^  y^  +  16  y*. 
9.  Sum  the  series : 

(i.)    —  3  —  2  —  1  ••.  to  n  terms, 
(ii.)       1  —  J  +  J^  •••  to  infinity. 

10.  A  train  72  yards  long  passed  another  train  60  yards  long, 
which  was  going  in  the  opposite  direction  on  a  parallel  line  of  rails, 
in  4  seconds.  Had  the  first-mentioned  train  been  travelling  at 
twice  its  actual  speed,  the  trains  would  have  passed  each  other  in 
3  seconds.  Find  the  number  of  miles  per  hour  at  which  the  trains 
were  travelling. 


408  MISCELLANEOUS   EXAMPLES   VL 

P.    1.   Simplify 
(x  +  y  +  zy-{-  x  +  y  +  zy+  (x  -  y  +  zy-  (x  +  y-  z)\ 

2.  Divide  a^  -  3  a^fe  +  3  a62  _  fts  _  ^3  by  a-b-  c. 

3.  Prove  that  if  x  =  a  -\-  d,  y  =  b  +  d,  z  =  c  +  d, 

then  x^  +  y^  +  z^  —  yz  —  zx  —  xy  =  a^  +  b^  +  d^  —  be  —  ca  —  ab. 


4. 

Simplify 

1                2a 

1 

a-2x     4x^- 

a'^     a 

;  +  2ic 

5. 

.   Solve  the  equations : 

(i.)  2x-^4:y-Sz 

=  22 

4x  —  2y  +  bz 

=  18 

■• 

5ic+    y-2z 

=  14, 

....  a;  -  1      a;  -  2  . 

X  — 

5     X- 

-6 

'^x-2     ic-3      x-6      ic-7 
(iii.)  ?i±l  +  ^-I=^  =  ^    1 
x2  +  ?/2  ^  90  J 

6.  Show  that,  if  any  integer  be  put  for  x  in  the  expression 

x6  _  4x5  +  14x4  -  32x3  +  49x2  -  60x  +  36, 
the  result  will  be  a  square  number. 

7.  Keduce  to  the  best  form  for  numerical  cal- 

V(12-V140) 
culations,  and  find  its  value  to  4  places  of  decimals. 

8.  Show  that  the  ratio  a  —  x :  a  +  x  is  greater  or  less  than  the 
ratio  a2  —  x2 :  a2  +  x2,  according  as  the  ratio  x  :  a  is  greater  or  less 
than  unity. 

9.  If  _  +  i  varies  inversely  as  x  +  y,  then  x^  +  y^  varies  as  xy. 

X     y 

10.    If  a,  b,  and  c  are  the  sums  of  n,  2  n,  and  3  n  terms  respec- 
tively of  a  geometric  progression,  then  a2  +  62  =  a(&  +  c). 


INEQUALITIES.  409 


CHAPTER   XXVII.  "^ 


Inequalities. 


276.  The  statement  p  —  q  is  positive  is  expressed  alge- 
braically hyp  —  q  >  0,  and  the  statement  p  —  q  is  negative 
is  similarly  expressed  hj  p  —  q  <  0.  With  this  notation 
the  definition  of  an  algebraic  inequality  becomes 

p>q,iip-q>Oj  [Art.  47.] 

and  it  immediately  follows  that 

p<q,iip  —  q<0. 

For  example,  since  —  3— (—  5)  =  2>0, 

.-.  -  3  >  -  5,  and  since  -9-(-6)=-4<0, 
...   _9<_6. 

This  definition  necessarily  involves  the  converse  propo- 
sitions that 

p  -  g  >  0,  if  p  >  9, 

^  _  g  <  0,  it  p<q. 

We  shall  call  either  of  these  two  forms  of  inequalities 
the  reverse  of  the  other. 
The  equational  statement 

p  =  q,  it  p-q  =  0 

is  the  transitional  form  through  which  either  of  these 
inequalities  passes  into  its  reverse. 


410  mEQUALITIES. 

277.  The  simpler  properties  of  inequalities  are  direct 
consequences  of  the  foregoing  definition. 

I.  Every  inequality  can  he  expressed  either  in  the  form 
p  >  0,  or  in  that  of  its  reverse  p  <  0 ;  —  an  obvious  conse- 
quence of  the  definition  of  an  inequality. 

II.  If  p^q  and  q  >  r,  then  p>r. 
For,  {p  —  q)-{-{q  —  r)  =p  —  r  >  0. 

Cor.  Ifp>q>r>S'-'>z,  then  p  >  z. 

III.  Any  term  may  be  transferred  from  one  side  of  an 
inequality  to  the  other,  provided  its  sign  be  changed. 

For,  ii  p  -\-  r  >  q,  then  by  definition  {p  -\-  r)  —  q  >  0, 
that  is,  jp  —  (Q'  —  ^)  >  0,  and  therefore  p  >  q  —  r.  Sim- 
ilarly, if  p  >  g  V  Sj  it  may  be  shown  that  p  —  s  >  q. 

IV.  The  signs  of  all  the  terms  of  an  inequality  may'  be 
changed,  provided  the  symbol  of  inequality  be  reversed. 

For,      if  m-\-p  —  r'>n-{-q  —  s,  then  by  III., 
—  n  —  g4-s>— wi  —p  -f  r, 
that  is,  —m—p  +  r<—n  —  q-{-s. 

278.  The  terms  of  an  inequality  may  be  combined  with 
the  terms  of  an  equation  in  accordance  with  the  follow- 
ing laws : 

V.  If  p>  q  and  r  =  s,  then  p±r'>q±s. 

For,  (p±r)  —  {q±s)  =  {p  —  q)±{r  —  s)=p-q=Si, 
positive  quantity.  ^,^ 

VI.  If  p  >  q  and  r  =  s,  then 

pr  >  qs  and  p/r  >  q/s,  if  r  and  s  be  positive ; 
pr  <  qs  and  p/r  <  q/s,  if  r  and  s  be  negative. 


(^ 


INEQUALITIES.  411 

For,  since  t=s,  and  p  —  g  is  positive  by  hypothesis, 

.'.  pr-gs,  p/r-q/s  =  {p-q)r,   {p-q)/r; 

whence  it  follows  that  pr  —  qs  and  p/r  —  q/s  are  positive 

when  r  and  s  are  positive,  but  negative  when  r  and  s  are 

negative. 

CoR.  1.  If  q  be  positive^  and  p  >  qr,  and  r  >  s,  then 
p  >  qs. 

Cor.  2.  If  the  terms  of  an  inequality  be  fractional,  they 
can  be  made  integral  by  the  proper  multiplications. 

Thus,  from  f  >  f  we  derive  5  x  5  >  4  x  6,  and  from >  — ; 

weobtain  (-4)x(-6)<5  X  5.  ^        ~^ 

279.  The  terms  of  two  or  more  inequalities  may  be 
associated  with  one  another,  but  under  certain  important 
limitations. 

VII.  If  Pi>qi,P2>  72,  Ps  >  ^3,  ••'  Pn  >  Qny 

then    ^i-i-p2+P3+---  +P„>  71  +  92  +  93+ •••+9n. 

For,  since  {pi  -  qi),{p2  -q'i),"'{pn-  Qn)  are  all  posi- 
tive and 

{Pi  +P2  +  —Pn)  -  (7i  +  72  +  •••  +7«) 

=  (i>i-7i)+-+(P«-g«), 

so  is  (pi+PaH hi>„)-(7i  +  72H 1-7„)  also  posi- 

tive,  and  therefore 

Pi  +P2  +  —  +i>n  >  7l  +  72  +  •••  +  7«- 

From  Pi  >  qi  andpa  >  72  we  cannot  infer  p^—  i?2>7i— 72- 
For  example,  8  >  7  and  5  >  2,  but  8  —  5  <  7  —  2. 

VIII.  If  Pi>quP2>  72,  Ps  >  73,  '•'Pn>  qny 

then  PiPiPi'-'Pn  >  7i7273---  7n, 

provided  all  the  quantities  be  positive. 


412  INEQUALITIES. 

For,  since  p^  >  qi  and  paPs  '-'Pn  is  positive, 

•••    PlP2"'Pn>qiP2P3"'Pny 

and  since  p2  >  ^2  ^-nd  QiPs-'-Pn  is  positive, 

.-.    PiP2"'Pn>qi<l2P3"'Pn, 

and  proceeding  in  this  way  until  all  the  p's  on  the  right 
are  replaced  by  the  corresponding  g's,  we  have  finally 

PlP2"'Pn>qiq2'"qn- 

From  pi  >  gi  and  ^2  >  9^2  we  cannot  infer  Pi/p2  >  Q'i/5'2- 
For  example,  8  >  6  and  4  >  2,  but  8/4  <  6/2. 

Cor.  J/*!)  and  q  be  positive  and  n  z=  a  positive  integer^ 
and  if  p  >  g,  then  p"  >  q"". 

IX.  i/*  j9  a?id  g  he  positive,  n=  a  positive  integer,  and 
^\/n^  gi/n^  (jlenote  the  positive  real  n*^  roots  of  p,  q ;  and  if 
p>q,  then  p^^""  >  q^^\ 

For,  if  this  be  not  so,  then  p^^""  <  or  =  g^/" ;  but  since 
p^/"  and  g^/"  are  both  real  and  positive,  fromp^/"  <  or  =  g^/" 
it  follows,  by  the  corollary  of  VIII,  that  (p^/")"<or 
__  (gV»)»i^  whence  p<  or  =  g,  which  contradicts  the 
hypothesis  p>q. 

CoR.  Ifp  and  q  he  positive  and  m,n  =  positive  integers, 
and  if  p>  q,  then  p"*/**  >  g«/";  provided,  if  n  he  not  1, 
only  the  positive  real  n'*  roots  he  taken. 

280.  The  following  examples  illustrate  some  of  the 
uses  of  the  method  of  inequalities  : 

Ex.  1.  The  sum  of  any  real  positive  quantity  and  its  reciprocal 
is  greater  than  2. 

Let  the  quantity  be  a,  and  let  a  ^/^  denote  its  positive  square  root. 
Then  (aV2  -  l/aV2)2  =  «_2  +  l/a>0; 

.-.  a-f  l/a>2.  [byllL] 


INEQUALITIES.  413 

Ex.  2.   For  what  values  of  xisx^  —  4x  +  3>  —  1? 
Completing  the  square  for  x,  we  have 

a;2  _  4x  +  3  =  (x  -  2)2  -  4  +  3. 
Hence,  the  given  inequality  is  satisfied  if 
(a;  _  2)2  -  1  >  -  1, 
that  is,  for  all  real  values  of  x  except  the  value  a;  =  2. 

Ex.  3.   For  what  values  of  x  is  (2  x  -  l)/(x  +  2)  >,  =,  or  <  1  ? 

Let  /=(2x-l)/(x  +  2). 

We  have  />  or  <  1, 

according  as         (2  x  -  1)  /  (x  +  2)  -  1  >  or  <  0,  [by  III.  ] 

according  as  {(2  x  -  1)  -  (x  +  2)}/  (x  +  2)  >  or  <  0, 
according  as  (x  -  3)  /  (x  +  2)  >  or  <  0, 

according  as  (x  -  3)  (x  +  2)  >  or  <  0  ;  [by  VI.] 

.•:/>or<0, 
according  as  x  >  or  <  3,  if  x  be  positive, 

or,  according  as   —  x  >  or  <  2,  if  x  be  negative  ; 
and  /=  1,   if  x  =  3. 

Ex.  4.  The  arithmetic  mean  of  any  two  positive  quantities  is 
greater  than  their  geometric  mean. 

If  the  two  quantities  be  a  and  6,  the  arithmetic  mean  is  \{a  -\-  b), 
the  geometric  mean  is  VCaft),  and  we  have  to  prove  that 

'    K«  +  ^)  >  V(a&). 
Since  (a  -  6)2  >  0, 

.-.  (a-6)2  +  4rt6>4a6,  [by  V.] 

that  is,  (a  +  6)2  >  4  a&  ; 

.-.  a  +  6>2v(a6),  [by  IX.] 

and  dividing  by  2  completes  the  proof. 

This  example  is  a  particular  case  of  the  following  : 

Ex.  6.  If  X,  y,  z,  '•'  the  n  real  quantities, 
(n-l)Sx2>2Sxy. 


414  INEQUALITIES. 

The  student  will  have  no  difficulty  in  verifying  the  indentity, 

by  comparing  the  like  terms  of  its  two  members.     But  S(x  —  ?/)2 
is  essentially  positive,  and  hence, 

(w- l)Sx2-2Sx«/>0, 
unless  oc  =  y  =  z  =  '••  =  t,   for  which  S(x  —  yy  =  0. 
.-.   (ri-l)2x2>2Sx?/. 
We  have  here  used  the  S  notation,  as  explained  in  Art.  152. 
The  student  should  note  its  conciseness  and  its  great  utility  in 
examples  of  this  kind. 

281.  The  theory  of  simultaneous  inequalities,  involv- 
ing two  or  more  unknown  quantities,  is  most  clearly  pre- 
sented with  the  aid  of  certain  geometrical  constructions, 
which  are  not  available  to  the  student  who  is  not  familiar 
with  at  least  the  elements  of  analytic  geometry.  But 
the  ordinary  process  of  elimination  by  addition  can  be 
applied  to  some  of  the  simpler  examples  of  this  class. 

It  must  be  remembered  that  multiplications,  with  the 
corresponding  terms  of  two  inequalities,  can  be  per- 
formed only  with  limitation,  and  subtractions  and  divis- 
ions not  at  all.     [Art.  279.] 

Ex.  1.   Resolve  the  simultaneous  inequalities 

2xf?/-6>0,    -3ic  +  2?/  +  6>0,   x-y  -]-l>0. 
We  apply  the  ordinary  process  of  elimination  by  addition. 
From  the  first  and  second  inequalities  we  obtain 
6x  +  Sy  -1S>0 
-6x  +  4y  +  12>0 

Ty-    6>0,    .-.  2/>f. 
From  the  first  and  third 

2x+y-e>0 
x-y+l>0 

3  X  -  5  >  0,   .  •.  X  >  |. 


INEQUALITIES.  415 

From  the  second  and  third 

-3x  +  2y  +  6>0 
2x-2y  +  2>0 

-x  +  8>0,   .-.  x<S. 
From  the  second  and  third 

Sx-2y  -6<0 
.  -3a;  +  3y-3<0 

y-9<0,  .'.  y<9. 
Thus,  the  values  of  x  and  y  are  confined  within  the  following 

limits : 

8>a;>|,   9>2/>f. 

Observe  that  if  x  and  y  are  to  be  limited  to  finite  values,  the 
inequalities  must  have  both  positive  and  negative  terms. 

Ex.  2.    Resolve  the  simultaneous  inequalities 
bx  +  ay  —  ab>  0,   xy  <  0. 
Reverse  the  first  inequality,  changing  its  signs  for  this  purpose, 
and  then  add  the  two  inequalities  together.     We  obtain 

xy  —  bx  —  ay  -{-  ab  <,  0, 
that  is,  (x  —  a)  (y  —  6)  <  0. 

.'.  x>a,  y  <b,   or  x<a,  y>b. 

If  a  and  b  are  both  positive,  since  x  and  y  are  of  opposite  sense, 
either  x  >  a,  y  <  0,  or  y  >  6,  x  <  0. 

EXAMPLES  LXXX. 

Determine  the  range  of  values  of  x  that  satisfy  each  of  the  fol- 
lowing inequalities : 

1.  (x-5)(x-2)>(a;-l)(a;-3). 

2.  a;2-4a;-i-3>,  =,  <0. 
>     Z.   x^-Sx+  10  >,  =,  <0. 

4.    (x  +  5)/(x-3)>  or  <  1. 

6.    (2x-3)/(x+3)>,  =,  or  <1. 

6.    (3x-4)/(2x-l)>,  =,  or  <1. 


416  INEQUALITIES. 

7.  Compare  the  values  of  -^(x  +  1)  +  ^/{'X,  —  1)  and  V(2^)  for 
real  values  of  x  and  positive  values  of  the  square  roots. 

8.  Resolve  the  simultaneous  inequalities 

a;+22/-4>0,   -3x  +  22/  +  6>0,  x-2/>0. 

9.  Resolve  the  simultaneous  inequalities 

x-i/  +  3>0,  2a;-4?/  +  8<0,  x+by-b<0. 

10.    Determine  the  values  of  x  and  y  that  make 

(x?/  +  l)/(x +  ?/)>,=,  or  <1. 

Discuss  the  cases  oj  +  ?/  >  0  and  x  +  y  <  0. 

.     11.    If  X,  y,  z  be  real,  prove  that 

Sx8>,  =,  or  <.Sxyz,  according  as  Sx>,  =,  or  <0. 

Use  the  result  of  Ex.  5,  Art.  280,  and  the  identity  of  Art.  164. 
[Chrystal,  Algebra,  Vol.  II.,  p.  40.] 

12.  If  X,  y,  z  be  real  and  not  all  equal,  prove  that 

(Sx)8>  27x^2;  and  <9Sx8. 

13.  If  a  >  6,  show  that 

{X  +  a)/^(ofi  +  a'^)>,  =,  or  <(x  +  &)/ V(«^  +  &^)» 
according  as  x>,  =,  or  <v'(a&). 

14.  Prove  that,  if  a,  6,  c,  c?  be  all  positive  and 

a/h  =  b/c  =  c/d, 

then  (a-d)>3(6-c). 

16.   Prove  that,  in  a  geometrical  progression  having  an  odd 
number  of  positive  terms, 

(sww  of  all  terms)'^ (number  of  all  terms)  x  {middle  term). 


LIMITS.      LNDETERjynNATB  FORMS.  417 


CHAPTER  XXVIII. 

Limits.    Indeterminate  Forms. 

282.  Functional  Nomenclature.  Any  expression  which 
involves  x  in  such  a  manner  as  to  undergo  a  change 
of  value  whenever  x  changes  is  called  a  function  of  x. 
If  two  or  more  letters,  as  x,  y,  z,  •  •  •  are  involved  in  it, 
the  expression  may  be  regarded  as  a  function  of  the  sev- 
eral letters  x,  y,  z,"-. 

These  letters,  through  whose  values  or  changes  the 
function  is  itself  determined  or  changed  in  value,  are 
called  the  variables,  and  all  other  letters  (including  mere 
numbers)  are  then  considered,  for  the  time  being,  as  not 
susceptible  of  change,  and  are  called  constants. 

If  the  function  does  not  contain,  or  can  be  so  reduced 
as  not  to  contain,  fractions  whose  denominators  involve 
^he  variables,  it  is  said  to  be  integral ;  but  if  such  denom- 
inators are  an  indispensable  part  of  the  function,  it  is 
said  to  be  fractional. 

The  term  rational,  or  irrational,  is  applied  to  the  function 
according  as  its  variables  do,  or  do  not,  appear  every- 
where in  it  in  the  rational  form. 

The  following  examples  will  sufficiently  illustrate  this 
functional  nomenclature : 

Ex.  1.    (i.)  axy  +  bx  +  cy  -\-  d  is  a  rational-integral  function  of  x 
and  y  of  the  second  degree. 
2d 


418  LIMITS.      INDETEKMINATE   FORMS. 

(ii.)  ax^-\-  hxy  +  cy'^  is  a  rational  integral  homogeneous  function 
of  X  and  y  of  the  second  degree. 

(iii.)   (ax  +  by  -{-  c) / (l  +  z)  is  a  rational  fractional  function  of 
X,  y,  and  z.    It  is,  however,  a  rational  integral  function  of  x  and  y. 


(iv.)  axy  +  bVx  +  y  is  an  irrational  function  of  x  and  y. 

(v. )  x^  +  y^  +  z^  -\-  xyz  is  a  rational  symmetrical  function  of  x, 
y,  and  z.    [Art.  150.] 

Ex.  2.  Describe  the  following  functions : 

(i.)  ax^  +  6?/^  +  Scxyz. 

(ii.)  ax  +  6?/  +  3c(xy0)i/8. 

(iii.)  (ax  +  by  +  c) / Vr+~z. 

(iv.)  ax2  +  2  6x  +  c. 

(v.)  a(x^-\-if)/Vxy. 

In  all  of  the  above  examples  the  letters  x,  y,  z  ave  regarded  as 
variables,  a,  6,  c,  (2  as  constants ;  and,  in  general,  constants  are 
denoted  by  the  letters  of  the  first  half  of  the  alphabet,  variables  by 
the  letters  of  the  second  half. 

283.  It  is  important  to  have  the  means  of  using  the 
ex^Tessions  function  of  x,  function  ofx,  y,  etc.,  in  mathe- 
matical formulae,  and  for  this  purpose  the  abbreviations 
f{x),  f(x,  y),  etc.,  are  employed. 

Thus,  such  an  expression  as  ax^  +  bxy  -\-  cy^  -\-  d  is  briefly  repre- 
sented by  /(x,  ?/) .     And  if 

/(x,  y)  =  ax^  +  bxy  +  cy'^  +  d, 

and  we  assign  x  =  1,  y  =  2,  then  we  write 

/(I,  2)=a.l  +  6.1.2  +  c-22+d 

=  a  +  26  +  4c  +  (^; 

similarly  f(l,  0)=  a  -h  d. 


LIMITS.      INDETERMINATE   FORMS.  419 

284.  A  function  of  x  is  continuous,  or  varies  continuously, 
if  its  change,  due  to  an  arbitrarily  small  *  change  in  x, 
is  also  arbitrarily  small. 

Thus,  f{x)  is  continuous,  at  the  value  x,  if  the  difference 
f(x  +  h)  —  f{x)  may  be  made  arbitrarily  small  by  making  h  arbi- 
trarily small. 

A  function  is  discontinuous,  at  any  value  of  x  for  which 
it  does  not  fulfil  the  foregoing  condition. 

285.  Definition  of  a  Limit.  If  by  making  x  approach 
the  fixed  value  a,  in  such  a  way  that  x  —  a  becomes  arbi- 
trarily small,  we  make  f{x)  approach  as  near  as  we  please 
to  the  fixed  value  Z,  then  I  is  said  to  be  the  limit  of  f(x) 
as  X  approaches  a.     We  express  this  briefly  by  writing 

It  will  be  convenient  to  use  the  sign  =  as  an  abbre- 
viation for  the  expression  approaches  as  a  limit. 

The  following  examples  will  help  to  make  the  mean- 
ing of  this  definition  clear. 

Ex.  1.    What  is  the  value  of  (x^  -  a^)/(x  -  a)  when  x  =  a? 

If  a  be  substituted  for  x  in  (x^  —  a^)/{x  —  a),  the  result  is  0/0, 
an  expression  which  cannot  be  interpreted  as  an  algebraic  opera- 
tion. There  is  therefore  no  proper  answer  to  the  question  as 
above  formulated.  But  expressed  in  the  form,  what  limit  does 
(x^  —  a^) / (x  —  a)  approach  when  x  =  a,  the  question  can  be 
answered  in  the  language  of  limits  as  follows  : 

Since  (x'^  —  a^) / {x  —  a)  =  {x  -\-  a){x  —  a) / (x  —  d)=  x  +  a, 
for  all  values  of  x  not  absolutely  equal  to  a,  we  see  at  once  that 

*  Smaller  than  a  previously  assigned  quantity  whose  value  we 
may  make  as  small  as  we  please. 


420  LIMITS.      INDETERMINATE   FORMS. 

when  X  differs  arbitrarily  little  from  a,  {^  —  a*^)  /  (x  —  a)  differs 
arbitrarily  little  from  2  a ; 

,                                       limit  a;2-a2  _  ^ 
hence,  ^      ^ =  z  a. 

Ex.  2.   Find  the  sum  to  infinity  of  the  series 


By  the  ordinary  process  of  division  we  have 

^^1^  =  X  +  a:2  +  x3  +  ...  +  x"-i, 

I  —X 

X  x^ 


that  is,  x  +  x2+ ...  +  a:'»-i  = 

1  —  X       1  —  X 

and  this  formula  is  true  for  all  finite  values  of  x. 
Putting  X  =  I,  we  have 

If  the  number  of  terms  of  the  series  be  now  increased  beyond 
finite  range,  l/2'*-i  =  0,  and  therefore  ' 


n  =  CO  \^2     4  2«-v 


This  series  was  discussed  in  Art.  266,  and  its  limit  was  there 
determined  from  geometrical  considerations. 

286.  If,  when  the  variable  approaches  a  limit,  the 
function  exceeds  every  finite  value  however  large,  we 
extend  the  language  of  limits  to  this  case  and  say  that 
the  function  has  the  limit  oo .  This  use  of  the  term 
limit  must  be  understood  to  be  exceptional. 

For  example, 

limit      x^  +  X  +  2       _  limit  a;  +  2  _  ^^ 

X=lx3-x2-X+l        X-ly?-  -\ 

The  value  0  may  obviously  occur  as  a  limit. 

^  1  limit  x^-\  _  r. 

For  example,  ^  ^  ^  -^^  -  ^' 


LIMITS.      INDETERMINATE   FORMS.  421 


INDETERMINATE  FORMS. 

287.  Certain  combinations  of  0  and  oo  produce  what 
are  known  as  indeterminate  forms.  Such  a  form  made 
its  appearance  in  Ex.  1  of  Art.  285,  where  the  substi- 
tution of  a  for  X  in  {x^  —  o?)  /  {x  —  a)  resulted  in  the 
illusory  fraction  0/0.  The  evaluation  of  indeterminate 
forms  is  essentially  an  application  of  the  method  of 
limits.  The  simpler  forms  are  0/0,  oo/oo,  0  x  co,  oo  —  cc. 
The  following  example^  will  show  how  they  may  arise 
and  exemplify  the  method  of  their  evaluation. 

Ex.  1.    (1  -  x5)/(l  -  a;-2)  becomes  0/0  when  x  =  \.     But 

l-x5_l-t-a;  +  a;^  +  a^  +  a^    1  -a;. 
I_x2~  1  +  a;  '  \-z' 

limit  \  —  x^  _  5 
•  •  X  =  1  1  _  x'-i  ~  2' 

Ex.  2.    (2  ic8  +  3  xS)  /  (3  x2  +  5  x*)  becomes  0/0  when  a;  =  0.  But 
2x»-|-3x5_2a;  +  3a^    a^. 
3x2  +  6x4      3  +  6x8  '  x^' 

.    limit  2x8  +  3x5_0_Q 
•  *  «  =  0  3x2  +  5x4     3       • 

Ex.  3.  (a  +  &'*)/(c  +  6")  becomes  qo/qo  when  6  >  1  and  n  =  00. 
If  numerator  and  denominator  be  multiphed  by  6-«,  then 

Umit    g  +  ft"_   limit   a&-»  +  1  _  ^ 
n  =  Qo  c  +  6'»~w  =  Qo  c^-**  +  1  ~   *^ 

If?)<l,  say  b  =  l/k,  where  A;>1,  then 

(a  +  &")  /  (c  +  &")  =  (a  +  A:-")  /  (c  +  A:-«), 

and  ^^"^'^    «  +  ^'"  =  «. 

n  =  Qo   c  +  A;-«      c 

If  6  =  1,  no  inference  can  as  yet  be  drawn. 


422  LIMITS.      INDETERMINATE   FORMS. 

Ex.  4.   The  function  (x  -  2)/(\/2x  -  2)  assumes  the  form  0/0 
when  X  =  2,  or  the  form  co/oo  when  x  =  co. 
In  the  first  case  we  have 

limit        X  —  2       _  limit   y/x  +  v/2  _  „ 
a^  =  2  ^(2 x)-2~^  =  2"^2       ~* 

In  the  second  case 

limit        X  — 2      _  limit         1  -  2/x       _1_ 

a^  =  '»   V(2a^)-2~^  =  °o  V(2A)-2/x"0~** 

Ex.  5.   The  function 


•^^^>  =  (^+^)*^'  +  ^-2> 


assumes  the  form  oo  x  0  when  x  =  1.    But 

1  _L     «     —^  +  a  —  1 
X  —  1  X  —  1 

and  X2  +  X-2  =:(x  +  2)(x- 1). 

.    limit    -,  X      limit  ,     ,  i\/^  ,  oa     q^ 

•••  a.^i/(a;)  =  ^^l(a;  +  a-l)(x  +  2)=3a. 

Ex.  6.    The  function 

/(x)=a/(l+x)-2&/(l-a;2) 
assumes  the  form  go  —  go  when  x  =  —  1.    But 

a/(l  +  a;)=a(l-a;)/(l-x2), 


and  therefore 


/w^^^^ff^- 


(i.)  Now,  in  whatever  manner  x  attains  the  value  —  1,  the  numer- 
ator of  this  fraction  assumes  the  unique  value  2  a  —  2  &,  and  the 
denominator  becomes  zero.  But  1  —  x:^  is  and  remains  positive  if 
x2  =  1  through  increasing  values  a  little  less  than  1,  or  it  is  and 
remains  negative  if  x^  =  1  through  decreasing  values  a  little  greater 
than  1.    In  the  former  case /(x)  =  +  2(a  —  6)  x  co,  and  if  a  >  6, 

limit    ^/  N 


LIMITS.      INDETERMINATE   FORMS.  423 

In  the  latter  case/(x)  =  —  2(a  —  6)  x  oo, 
and  if  a  >  Z>, 

The  function  has  two  branches,  one  for  values  of  cc  less  than 
—  1,  the  other  for  values  of  x  greater  than  —1,  and  each  branch' 
has  its  own  limit  at  the  value  a;  =  —  1. 

(ii.)  If  &  =  a,  the  function  becomes 

/•/3.N^-a(l  +  a;)^     a      .  1  +  a; 
^^  ^         l_x'-i         x-\     1  +  x 

As  an  indeterminate  form  it  has  been  changed  to  0/0,  and  its 
limit,  for  ic  =  —  1,  is  obviously  —\a. 

288.  The  fundamental  indeterminate  form  is  0/0,  to 
which  all  the  others  may  be  reduced.  This  property 
results  from  the  definition  of  infinity,  as  given  in  Art. 
168,  in  accordance  with  which,  if  a  and  h  be  any  finite 
quantities,  the  symbol  oo  may  be  replaced  ad  libitum  by 
any  such  value  as  a/0  or  6/0. 

Hence  we  may  write 

QO_a  /b__a     0_0-a_0 

a      b     0-a  —  O'b     0 

00  —  oo  := =  =  — 

0      0  0  0 

Ex.  1.  The  function  a/(l  -x)-2  a/ (I  -  x^)  becomes  oo  -  oo 
when  x  =  l.     But 

a 2o__  a(l  -\-  x)—2a 

1  -  ic     1  -  x2  ~        1  -  x2       ' 


and  in  this  form  it  becomes  0/0  when  a;  =  1. 


424  LIMITS.      INDETERMINATE   FORMS. 

Ex.  2.  The  function  (1  -  a'*)aV«,  where  a  >  1,  becomes  0  x  oo 
when  n  =  +  0.    But 

and  in  this  form  it  becomes  0/0  when  w  =  +  0. 

289.  That  any  quantity  whatever  may  appear  as  the 
result  in  the  evaluation  of  any  of  the  indeterminate 
forms  thus  far  considered  is  at  once  evident  from  the 
consideration  that  the  one  form  0/0  may  be  taken  as  the 
general  representative  of  the  entire  set  of  four  forms, 
and  that,  a  being  an  arbitrary  finite  quantity, 

a  X  0  =  0,  0x0  =  0,  and  O/oo  =0; 

for,  from  these  three  equations  we  derive,  by  the  proper 
multiplications  and  divisions, 

a  =  0/0,   0  =  0/0,  and  ao  =  0/0, 

respectively.      The   multiplications   and   divisions   here 
performed  are,  of  course,  merely  symbolic. 

The  important  fact  to  be  remembered,  in  this  connec- 
tion, is  that  the  indeterminate  form  gives  no  information 
concerning  the  critical  value  of  the  function  that  gives 
rise  to  it. 

EXAMPLES  LXXXI. 

Evaluate  the  limits  of  the  following  functions  for  the  values  of 
X  indicated,  taking  account  of  positive  square  roots  only. 

1.  (x2  -  X  -  2) /(x2  +  a;  -  6)  when  x  =  2. 

2.  (x3  -x^-x  +  l)/(x5  +  x^  -5x  +  S)  when  x  =  1. 

3.  (ic«  -  a")  /{x  —  a)  when  x^a. 

4.  (a:3/2  _  «3/2)  /  (^1/2  _  ^1/2)  when  a;  =  a. 

5.  (V^c- l)/>/(a:  -  1)  whena;  =  1. 


L 


LIMITS.      INDETERMINATE   FORMS.  425 

6.  {2  -  ^(x  +  3)}/(l  -  x)  when  x  =  1. 

7.  -—^ ^ — wheii?;  =  l. 

V(l-v'^)      VCl-^) 

8-    V^-^-VC^-l)  whenx  =  l.        ^QtXu<Hv^-v'wv\. 
(« -  1)2        ^a;  -  1  -J 

2  1 

9. when  x  =  2. 

2-x     2-  ^'Ix 

10.  {2  -  V(l  +  a;)}/{3  -  V3^}  when  x  =  3. 

1 1     ax*  +  6x2  +  ex  +  <Z 

11.  — ' =!—  when  x  =  00- 

It^  +  mx  +  w 

1  o      ax*  +  6a;  +  c      , 

12.  — - — ■ — ^ —  when  x  =  oo. 

Zx^  +  wx2  +  n 

18.    Solve  the  equation  +  ^{y?  _  9)  -  (x  -  9)  =  0.     [See  page 
248.] 

14.    Solve  the  equation  +  VC^^  +  2  ax)  -  (x  +  a)  =  0. 


i26  EXPONENTIALS  AND   LOGARITHMS. 


CHAPTER   XXIX. 

Exponentials  and  Logarithms. 

290.  The  inverse  operations  of  raising  to  powers  and 
extracting  of  roots,  considered  in  Chapter  XX.,  may  be 
regarded  as  presenting  themselves  through  the  medium 
of  the  equation 

in  the  two  inverse  problems  : 

(1)  Given  a  Jcriown  number  x  and  a  known  integer  k,  to 
find  y.     {Involution.) 

(2)  Given  a  known  number  y  and  a  known  integer  k, 
to  find  X.     {Evolution.) 

In  these  problems  the  exponent,  as  it  appears  either  in 
x^  or  in  y/*,  is  always  to  be  regarded,  either  as  an  integer, 
positive  or  negative,  or  as  the  reciprocal  of  such  an  inte- 
ger ;  the  case  where  it  is  a  fraction  whose  numerator  is 
not  unity  calling  for  the  application  of  the  processes  of 
involution  and  evolution  in  succession. 

Thus,  if  the  form  6^/^  presents  itself,  we  first  find  6^,  or 
216,  and  seek  the  fifth  roots  of  216 ;  or,  we  may  reverse 
the  order  of  this  work  and  get  first  the  fifth  roots  of  6 
and  then  cube  each  one  of  them. 

Involution  and  evolution,  therefore,  deal  primarily 
with  such  expressions  as 

2%   3^   8^^  etc., 
and  2V3,.  ^y\   lliA   etc. 


EXPONENTIALS   AND   LOGARITHMS.  427 

291.  If  now  the  form  of  the  expression  d"  be  changed 
to  }f,  for  the  purpose  of  indicating  that  the  exponent 
may  become  an  unknown  quantity  (the  variable),  the 
equation 

gives  rise  to  the  two  inverse  problems : 

(1)  Given  x  and  b,  any  known  numbers,  to  find  y. 

(2)  Given  y  and  b,  any  known  numbers,  to  find  x.  ^ 
The  operation  upon  b  which  produces  b'  is  called  expo- 
nentiation, b  is  called  the  base,  and  6*  (sometimes  also 
written  exp^a;)  is  called  the  exponential  of  x  with  respect 
to  the  base  b. 

The  inverse  operation  of  producing  x,  when  b  and  y  are 
given,  is  called  logarithmic  operation,  and  is  expressed  sym- 
bolically by  the  equation 

x  =  nog  y, 
in  which  *log  y  is  read  logarithm  of  y  with  respect  to  the 
base  b. 

It  is  an  immediate  consequence  of  this  definition  that 

EXPONENTIATION. 

292.  The  process  of  exponentiation  seeks  the  value  of 
expressions  of  the  form  b',  in  which  x  and  b  are  any 
real  numbers.f     But  in  thus  generalizing  the  combined 

*  A  third  case  may  also  present  itself  ;  namely,  given  x  and  y,to 
find  h.  But  this  is  identical  with  (1),  for  the  equation  h'  =  y  may 
be  replaced  by  6  =  yV*. 

t  In  the  most  general  interpretation,  exponentiation  also  allows 
h  and  x,  in  the  expression  6^,  to  have  imaginary  and  complex  values. 


428  EXPONENTIALS  AND  LOGAEITHMS. 

operations  of  involution  and  evolution,  it  so  restricts  (by 
arbitrary  definition)  the  interpretation  of  its  results  that 
to  an  unique  pair  of  values  of  x  and  h  there  corresponds 
one  and  only  one  value  of  If,  so  that  while  6^/*,  regarded 
as  an  example  in  evolution,  has  h  roots,  as  the  exponen- 
tial of  l/k  to  base  h  it  has  but  one  value.  The  expo- 
nential will  therefore  be  so  defined  as  to  embody  this 
restriction. 

As  already  indicated,  this  restriction  is  arbitrary,  but 
in  the  chapter  on  exponential  and  logarithmic  series,  a 
definite  algebraic  form  will  be  assigned  to  6'=,  namely,  a 
series  whose  terms  are  integral  powers  of  x  having  co- 
efiicients  that  involve  only  h  and  certain  numerical  frac- 
tions, and  in  this  form  the  uniqueness  of  the  value  of  Jf, 
for  a  given  value  of  x,  will  be  apparent. 

293.  In  defining  If  for  our  present  purposes  we  suppose 
that  h  is  real,  positive  and  greater  than  1,  and  should  x 
be  irrational,  we  replace  it  by  one  of  the  rational  frac- 
tions m/n,  or  (m  +  V)/n,^  which,  by  properly  choosing 
the  integers  m  and  n,  may  be  made  to  differ  from  x  by 
an  arbitrarily  small  quantity,  that  is,  by  a  quantity  that 
is  smaller  than  1/w,  where  n  is  as  large  as  we  choose  to 
make  it.     [See  Art.  249.]     The  values  of  h""  we  have  to 

But  the  discussion  of  these  more  general  cases  does  not  fall  within 
the  scope  of  this  elementary  treatise.  For  a  fuller  Interpretation 
of  exponentiation  and  logarithmic  operation  the  student  is  referred 
to  Chrystal's^ Z^e&ra,  Vol.  II.,  Chapter  XXIX.,  andtoStringham's 
Uniplauar  Algebra,  Chapter  III. 

*It  must  not  be  inferred  that  these  restrictions  are  necessary  in 
the  general  theory  of  exponentials.     [See  foot-note  to  Art.  292.] 


EXPONENTIALS   AND   LOGARITHMS.  429 

consider,  are  therefore  of  the  form  ft*^/",  in  which  m  and 
n  are  integers,  and  we  retain  only  the  real  positive  n^^ 
root. 

For  negative  values  of  x,  say  a;  =  —  m/n,  h'  becomes 
1/6'"/'*,  namely,  the  reciprocal  of  an  exponential  with 
positive  exponent. 

294.  The  Laws  of  Indices  as  applied  to  expressions  of 
the  form  6*,  under  the  limitations  above  imposed  upon 
the  values  of  h  and  x,  have  been  fully  explained  and 
justified  in  Chapters  XX.,  XXI.,  and  when  we  require 
their  application  to  exponentials  and  logarithms  we  may, 
therefore,  write  without  hesitation, 

6^6*6*..- =  6"+"+'+-, 
h'/h"  =  ¥-^, 
(p')y  =  {byy  =  h^, 
{ahC"'y  =  a'h'<f"'. 

295.  If  a  series  of  numbers  he  in  arithmetical  progres- 
sion, their  exponentials,  with  respect  to  any  base,  are  in 
geometrical  progression. 

This  proposition  follows  immediately  from  the  equa- 
tions 

6- =6-,  b'+^^b'b'', 


Here  d  is  the  common  difference  of  the  arithmetical 
progression 

X,   X  -\-d,   x-\-2d,"-x-{-  nd, 

and  ¥  is  the  common  ratio  of  the  geometrical  progression 

b',   b'b",   h^b'^r-b^b'^- 


430  EXPONENTIALS   AND   LOGARITHMS. 


LOGARITHMIC   OPERATION. 

296.  We  have  already  defined  the  logarithm  to  be  a 
value  of  X  obtained  from  the  equation  y.  =]f^  when  6  and 
y  are  given,  and  we  denote  this  value  of  x  by 

JB=?=^l0g2/. 

The  properties  of  logarithms  follow  immediately  from 
the  fundamental  laws  of  indices.  We  proceed  to  deduce 
them  and  to  apply  them  to  a  few  important  examples. 

297.  Properties  of  Logarithms. 

I.  Since  IP  =\  for  all  finite  values  of  6,  it  follows 
that  4og  1  =  0;  that  is  : 

The  logarithm  of  1  is  0,  whatever  the  base  may  he. 

II.  Since  h^  =  h  for  all  values  of  b,  it  follows  that 
*log  &  =  1 ;  that  is  : 

The  logarithm  of  any  number  with  respect  to  that  num- 
ber as  base  is  always  1. 

III.  li^logx=p  and  *log?/  =  g;  then,  by  definition, 

x=b^  and  y  z=b^-^ 

.'.  xxy  =  bPxb^  =  ¥'^^ ; 

.-.  4og(a;2/)=j9  +  Q'  =  *loga;4-nogy. 

Similarly  it  can  be  proved  that 

^\og{xyZ'")  =  ^\ogx-\-^\ogy  -\-^\ogz-\ . 

That  is :  The  logarithm  of  a  product  is  the  sum  of  the 
logarithms  of  its  factors. 


EXPONENTIALS   AND   LOGARITHMS.  431 

IV.  If4ogic=p  and  ^logy  =  q;  then,  by  definition, 

x=b^  and  y=¥', 

.-.  xjy=  6^/6«  =  6P-*; 

.\^\og{x/y)=p-q  =  nogx-'>\ogy.    . 

That  is :  The  logarithm  of  a  quotient  is  the  algebraic 
difference  of  the  logarithms  of  the  dividend  and  the  divisor. 

V.  If  ic  =  6^ ;  then  x""  =  bP""  for  all  values  of  m.    Hence 

*log af*  =  Hog bP^=p  xm  =  Hog x  x  m. 

That  is :  The  logarithm  of  ayiy  power  of  a  member  is 
the  product  of  the  logarithm  of  the  number  by  the  index 
of  the  power. 

VI.  Let  *log  x=p  and  *log  x  =  q. 
Then  '  a;  =  a^,   and  x  —  ¥. 
Hence  a^  =  6',  and  therefore 

a  =  6'/^  and  b^a^f"-, 
and,  by  definition  of  a  logarithm, 

Hog  a  =  q/p,  and  "log  b  =  p/q ; 

.-.  Hoga  X  "log 6  =  ?  X  -  =  1. 
P     <1 

Again,  from  q=p  x  Hog  a, 
we  have  *log  x  =  Hog  x  x  Hog  a. 

Hence :  TJie  logarithms  of  a  series  of  numbers  to  the 
base  b  are  found  by  multii^lying  the  logarithms  of  those 
numbers  to  the  base  a  by  the  constant  multiplier  Hog  a. 


432  EXPONENTIALS    AND   LOGARITHMS. 

VII.  Since  6+*  =  +  oo ,  and  6"*  =  0,  for  all  positive 
finite  values  of  h,  \h  >  1], 

4og(  +  oo)  =  +  oo,  and  4ogO  =  — oo. 

Hence :  In  every  system  of  logarithms  the  logarithm  of 
-j-oo  is  +00,  and  the  logarithm  of  0  is  -co. 

298.  The  logarithms  in  any  system  whose  base  is  real, 
positive  and  greater  than  1  [see  Art.  293]  have  the  fol- 
lowing properties. 

VIII.  The  logarithm  of  a  negative  number  cannot  he  a 
real  number. 

For,  if  b  be  positive,  and  a;  be  a  real  number,  6^  cannot 
be  negative ;  that  is,  x  cannot  be  the  logarithm  of  a  nega- 
tive number. 

IX.  Numbers  greater  than  1  have  positive  logarithms^ 
numbers  less  than  1  have  negative  logarithms. 

For,  if  b""  —  u  and  6  >  1,  then,  since  for  n  =  a  positive 
number  6"  >  1,  and  6~"  <  1, 

.-.  x>,  or  <0,  according  as  w  >,  or  <  1 ; 
that  is,  since  x  =  *log  u, 

4og  w  >,  or  <  0,  according  as  w  >,  or  <  1. 

Note.  —  In  a  system  whose  base  is  positive  and  less  than  1, 
numbers  greater  than  1  have  negative  logarithms,  and  numbers  less 
than  1  have  positive  logarithms.     For,  by  VI.,  Art.  297, 

^        Hog{l/a)         nog  a 

and  if  a  and  b  be  both  greater  than  1,  by  the  proposition  just 
established,  ^loga  is  positive,  and  *logw  is  positive  or  negative, 
according  as  u  is  greater  or  less  than  1. 


EXPONENTIALS  AND   LOGARITHMS.  433 

299.  If  a  series  of  numbers  be  in  geometrical  progression 
their  logarithms  are  in  arithmetical  progression. 

This  is  the  proposition  of  Art.  295,  stated  in  terms  of 
logarithms  instead  of  exponentials.  Thus,  the  loga- 
rithms of  the  several  terms  of  the  geometrical , progres- 
sion 

are  4og  y,  4og  y  -f-  4og  r,  •  •  •  'log  y  -{-n  4og  r. 

300.  The  following  examples  will  illustrate  this  part 
of  our  subject. 

Ex.  1.  Find  21og8,  nog  2,  wioglOOO,  and  ^log  ^100. 
8  =  23,  2  =  8^,  1000  =  103,  and  ^100  =  loi 
Hence  21og8  =  3,  81og2  =  ^,  loiog  1000  =  3,  and  loiog  ^100  =  f. 

Ex.  2.  Having  given  .ioiog2  =  .3010300  and  i<51og3  =  .4771213, 
findioiog40,  i^og  12,  ioiogl5,  and  loiog  ^880. 

By  the  laws  of  logarithmic  operation  we  have  : 
ioiog40  =  loiog  (4  X  10)=  ioiog2^  +  i^loglO 

=  2  X  .3010300  +  1  =  1.6020600. 

loiog  12  =  loiog  (22  X  3)  =  2  X  ioiog2  +  ioiog3 

=  2  X  .3010300  4-  .4771213 
=  1.0791813. 

ioiogl5  =  ioiog^i^^^=  wioglO  +  ioiog3  -  wiog2 

=  1  +  .4771213  -  .3010300 
.  =  1.17G0913. 

loiog  ^880  =  I  X  loiog  (10  X  25  X  32) 

=  K^°log  10  +  5  X  i'^log2  +  2  X  wiog3) 
=  1(1  +  1.5051500  +  .9542426) 
=  1.1531308. 
2e 


434  EXPONENTIALS   AND   LOGARITHMS. 

Ex.  3.    Solve  the  equation  a'^^+<'  =  d. 

Taking  the  logarithms  of  both  sides  of  the  equation,  with  respect 
to  base  a,  we  have 

6x  +  c  =  «log  d. 

.  '-logd-c 

•  h 

Ex.  4.    Solve  the  equation 

f(x)  =(2^  .  ai*-3  -  2*  \  a^-2  -  4  •  a^x-i  +  4  ==  o. 
The  left  member  of  the  equation  may  be  factored  thus : 
fix)  =  2*  .;a^-2  (a3x-i  _  1)  _4(a3^-i  -  1) 
=  (2«.a^-2-4)(a3x-i_i). 
Hence  2*  •  a'--  -4  =  0,  or  a^x-i  _  1  _  q  ; 

and,  therefore,  by  logarithmic  operation,  in  any  system, 
xlog2+  (ic  -  2)  log  a  -  log 22  =  0, 
{X  -  2)(log  2  +  log  a)  =  0,  a;  =  2, 
or  (3  X  —  1)  log  a  =  0,  x  =  \. 

EXAMPLES  LXXXII. 
Find: 

1.  loiogl.                       6.   loiogV-Ol.  9.  ^2iog2. 

2.  loiog  10000.               6.    loiog^lOO.  10.  ^logSv^. 

3.  loiog.Ol.                   7.   2log32.  11.  nog  3^3. 

4.  loiog.OOl.                 8.  nog  32.  12.  ^log  ^64. 

13.  Having  given  ^log  2  =  .3010300  and  i^log  3  =  .4771213,  find 
loiogf,  loiogOO,  ioiog4500,  and  ^log  ^25. 

14.  Given     loiog  3  =  .4771213     and    ^log  5  =  .6989700,     find 
i^log3.75,  ioiogl.28  and  wiog(33  x  'o^f2J). 

15.  Given  ^log  5  =  .6989700  and  i^log  6  =  .7781513,  find  ^  ,g  324, 
loiog  1.458,  and  ^log. 00432. 

16.  Given     I'^log  5  =  .6989700    and     loiog  7  =  .8450980,     find 
ioiogl.25,  ioiogl.28,  and  ^log  (23  x  1^/^^). 


EXPONENTIALS   AND   LOGARITHMS.  435 

"^      17.    Given  i^log  12  =  1.0791813   and  ^log  18  =  1.2552726,    find 
io;og8  and  ioiog9. 

18.  Given  ^log  24  =  1.3802113  and  wiog  36  =  1.5563026,  find 
loiog  72  and  i^log  ^432. 

19.  If  a,  6,  c  be  tlie  bases  of  any  three  systems  of  logarithms, 
p  .ve  that 

51oga:=M«.alogX. 

«log6 

20.  Prove  that,  in  a  series  of  logarithmic  systems,  whose  bases 
are  respectively  a,  &,  c,  etc., 

"log  a;  ^  ^loga;  _  ''logx  _  ^^^^ 
«logy     ^logy     «logy 

21.  Given  «log2  =  69315  and  «log  5  =  1.60944,  find  ioiog80, 
61og64,  and  2log  125. 

22.  Given  «log  5  =  1.60944  and  'log  3  =  1.09861,  find  6log27, 
81og25,  and  ^log  15. 

23.  If  loiog  30.6614  =  1.48659  and  ^log  3.1299  =  .49553,  what  is 
the  relation  between  30.6614  and  3.1299  ? 

24.  If  loiog  .3.0501  =  .4813141,  i^log  9.458055  =  .9758019  and 
I'^log  3.1009  =  .4914878,  what  is  the  relation  between  3.0501,  3.1009, 
and  9.458055? 

X      25.   Given  ioiog2  =  .301030,  find  x  from  the  equation  2*  =  10. 

26.  Given  ^log  2  =  .301030  and  ^log  5  =  .698970,  find  x  from 
the  equation  2=^-^  =  10*. 

27.  Solve  the  equation  2^  -  2*+^  -f  1  =  0. 

28.  Solve  the  equation  52*  —  5*+i  +  6  =  0,  obtaining  two  values 
of  X  in  terms  of  the  logarithms  of  2,  3,  and  5. 

>•'  29.    Solve  the  equation  a^~^  -  a^-^  —  a*  +  1  =  0. 

30.    Solve  the  equation  a'^^  -\-  b  ■  a''  —  c  =  0. 
^  31.   Solve  the  equation  2*  •  3^'-  3*  -  9  •  6*  +  9  =  0. 

32.  "l^olve  the  equation  3*'-2*+2  =10,  obtaining  two  values  of  x 
in  terms  of  the  logarithm  of  3  to  base  10. 

33.  Solve  the  simultaneous  equaii6'fi% 

2x+2y  -  10,    53y-i  =  25*+y, 
obtaining  x  and  y  in  terms  of  the  logarithm  of  2  to  base  10. 


436  NATURAL  LOGARITHMS. 


CHAPTER   XXX. 

Natural  Logarithms. 

301.  The  limit  which  the  ratio  {^logy'—^logy)/{y^—  y) 
approaches  when  2/'  and  y  both  approach  the  common 
limit  1  is  called  the  modulus  of  the  system  of  logarithms 
whose  base  is  6-     Let  y'  =  ry ;  then  this  ratio  becomes 

4og?*z/— 4og?/        4ogr 


ry-y  y{r  —  iy 

and  when  y  =  l  it  is  (4og  r)/{r  —  1).*  Hence  the  modu- 
lus may  be  defined  as  the  limit  which  (*logr)/(r  —  1) 
approaches  when  r  approaches  1 ;  or,  symbolically,  if  M 
denote  the  modulus, 

r  =  1 r  —  1 

When  the  modulus  is  1  the  corresponding  system  of  log- 
arithms is  called  the  natural  or  Napierian  system,t  and  the 
base  of  such  a  system  is  called  natural  or  Napierian  base.f 

*  This  modulus  is  otherwise  described  as  the  rate  of  change 
of  ^logr  at  the  instant  when  r  =  1.  [See  Stringham's  Uniplanar 
Algebra,  Art.  23.] 

t  The  logarithms  of  the  natural  system  must  not  be  confounded 
with  the  numbers  of  Napier's  original  tables,  which  were  not  cal- 
culated with  respect  to  a  base,  and  are  therefore  different  from  the 
so-called  Napierian  logarithms  of  modern  tables. 


NATURAL  LOGARITHMS.  4B7 


and  is  denoted  by  the  letter  e.    We  therefore  have,  by 
definition, 

limit  "logr*^^ 

r= 1 r— 1 

302.  From  this  definition  we  are  able  to  derive  a  for- 
mula by  means  of  which  an  approximate  value  of  the 
number  e  may  be  calculated,  and  this  being  done,  the 
modulus  corresponding  to  any  base  b  will  be  given  by 
the  formula 

limit 'log  ^ 


M= 


r=z  1  r  —  1 


=  '  =  1F^'1°S«'      [Art.297,VL] 

that  is,  since  limit  (4og  r)/(r  —  1)  =  1, 

3/=*loge  =  l/'log6. 

In  place  of  *log  we  shall  henceforth  use  the  shorter 
symbol  In,  made  up  of  the  initial  letters  of  logarithm 
and  of  natural  or  Napierian. 

303.   We  first  derive  a  formula  which  determines  e  as 
a  numerical  limit. 
For  this  purpose  let 

In  r  =  n,  that  is,  r  =  €", 

and  let  u/{e^  —  1)  =  h,  or  e"li  =  h-\-u. 

*  This  fraction  is  positive  whether  1  be  greater  or  less  than  1, 
for  the  logarithm  of  any  number  less  than  1  is  negative  [Art.  298]. 
Hence  r  may  approach  1  through  values  either  a  little  smaller  or 
a  little  greater  than  1. 


438  NATURAL  LOGARITHMS. 

Then  u  =  0  when  r  =  1,  and  /i  =  1  when  u  —  0,  and  we 
have  [Art.  301]  ^ 

limit    Inr        limit       u 


But  from  eVi  =  h  -{-  u  we  obtain 
.   e''  =  (l+i6//i)'^/"; 
and  hence,  since  h  =  l  when  u=0, 
limit 


=  limit  ^  =  1. 


e  = 


„  =  o(i +  «)■'"• 


We  have  here  supposed  u  to  approach  zero  from  the 
positive  side,  but  the  supposition  was  not  a  necessary- 
one.  If  we  assign  to  u  Si  negative  value,  say  —v,  the 
formula  for  the  value  of  e  becomes 

limit 
u  =  0 


e=  Zai^-v)-'-. 


304.  Now  it  may  be  shown  that  in  whatever  way  u  is 
made  to  approach  zero,  whether  positively  or  negatively, 
the  limit  of  (1  +  it)^/"  is  a  finite  number,  and  that  when 
u  passes  continuously  towards  zei'O,  (1  +  2^)^/"  diminishes, 
and  (1  —  w)^/"  increases,  continuously  towards  a  common 
limit ;  and  that  e  is  therefore  a  definite  number. 

For  a  formal  proof  of  these  statements,  which  is  not 
properly  within  the  scope  of  this  treatise,  we  refer  the 
student  to  Chrystal's  Algebra,  Vol.  II.,  Chapter  XXV., 
§  13,  and  only  present  here  a  few  numerical  verifications. 
By  ordinary  arithmetical  computation  we  easily  find : 


(1  +  1)^     =2.25. 

-i)-'     =4. 

(1  +  1)3    ^2.370. 

-i)-3     =3.5. 

(1+1)4    =2.44140625. 

(1- 

-^)-4      =3.1604903... 

{1  +  ^y    =2.48832. 

-1)-^     =3.0517578125. 

a  4.  _ij^)io  =  2.5937424601. 

-^y -10  =  2.867971990.. 

NATURAL  LOGARITHMS.  439 

Thus,  as  the  exponents  m  (1 -j- ?*)^/"  and  (1  — r^)"^/" 
are  made  numerically  larger,  the  two  series  of  numbers 
produced  by  these  formulae  slowly  approach  one  another, 
those  in  the  left  column  by  increase,  those  in  the  right 
column  by  decrease. 

We  can  infer  from  this  calculation  that  e  is  greater 
than  2.5  and  less  than  2.9.  None  of  the  decimal  figures, 
however,  in  any  of  these  numbers  are  as  yet  correct  for  the 
value  of  e,  and  further  extensions  of  this  method  become 
rapidly  tedious.*  We  therefore  do  not  use  the  formula 
(1  -h  m)^/"  for  obtaining  close  approximations  to  the  value 
of  e,  but  replace  it  by  a  rapidly  converging  series. 

305.  Such  a  series  may  be  obtained  by  expanding 
(1-j-  l/z)"  by  the  binomial  theorem,  on  the  supposition 
that  2  is  a  positive  integer  and  ultimately  becomes  infi- 
nitely large. t  In  this  investigation  we  shall  use  the 
convenient  symbol  n !  to  denote  the  product  of  the  first 
n  natural  numbers  ;  that  is, 

n !  =  1  •  2  •  3  •  •  •  to  71  factors. 

With  this  notation  it  is  obvious  that 

nl=  {n  —  1)\  n  =  (?i  —  2)!(n  —  l)n,  etc. 

By  the  binomial  theorem  we  have,  for  integral  values 
of  z  [Art.  205], 

♦  The  value  of  e,  correct  to  seven  places  of  decimals,  is  2.7182818, 
and  the  value  of  (1  +  .001)^ooo  gives  only  two  of  these  decimal 
figures.  In  fact,  the  first  eight  figures  in  the  computed  value  of 
(1  +  •001)«wo  are  2.7169239. 

t  The  demonstration  is  taken  from  Chrystal's  Algebra,  Vol.  II., 
Chapter  XXVIII.,  §  2. 


440  NATURAL  LOGARITHMS. 

\        zj  z  21       z^  3!  s^ 

,  z(g-l)...(2;-n  +  l)l    ,         ,1 
n!  2**  2;* 

that  is, 

(^       2^         ^2!  3!  ^ 


_^  (i-iA)-(i-»-iA)  _^  _g 

where 


^  ^(l-lA)...(l-nA)      (l-l/.)...(l-n+lA) 

(71  +  1)!  (n  +  2)!  ^     ' 

Let  n  have  any  /a;ec?  integral  value  and  let  z  be  an 
arbitrarily  large  integer,  at  least  larger  than  n.  The 
series  B^  will  then  terminate  ;  and  each  of  its  numera- 
tors will  be  less  than  1,  for 

1/z  <  2/z <3/z<"'  < n/z <"'<  (z  -  l)/z <  1. 
Hence 

r»  ^         1 I 1  I 1 I  4-  _ 

"     (n  -h  1) !      (n  +  2)  !      (n  +  3)  !  2 ! 

<_^^ 1 

(n  +  1)!  1-1/(71  +  2) 

If  now  in  the  second  of  the  above  series  for  (1  +  l/z)' 
we  put  2J  =  00,  each  of  its  numerators  becomes  1,  and  we 
have 

i^""!!  fi + -T=  1 + T + li + li +•••+-.+ -^^ 

z  =  Gc\        zJ  12!      31  ill 


NATURAL  LOGARITHMS. 


441 


where  Bn  satisfies  the  condition 
1 


Rn< 


1 


(n  +  l)!l-l/(n  +  2) 

Finally,  if  n  be  made  to  increase  beyond  finite  range, 
J?„  decreases  towards  the  limit  zero,  the  number  of  terms 
in  the  series  becomes  indefinitely  great,  and  we  have 

limit  j^i  ^  ly  _  1  +  1  ^  1  _^  ...  1  ^  ...  ^^  infinUum. 
z=ccy       zj  12!  r! 


306.  This  is  the  series  by  means  of  which  the  value 
of  e  is  calculated  to  as  great  a  degree  of  accuracy  as  is 
desired.  The  third  term  is  .5,  the  fourth  is  .5/3,  the 
fifth  is  this  last  quotient  divided  by  4,  and  so  on.  Set- 
ting down  these  results  in  succession  as  far  as  the  term 
1/10 !  we  have 

1  +  1  =  2. 


1/2!  = 

.5 

1/3!  = 

.1666667 

1/4!  = 

416667 

1/5!  = 

83333 

1/6!  = 

13889 

1/7!  = 

1984 

1/8!  = 

248 

1/9!  = 

27 

1/10 ! = 

3 

.*.  approximately,       e  =  2.7182818. 

This  result  is  an  approximate  value  of  the  series 

1 


l  +  l  +  i+i^.., 
^1^2!^3!^ 


+  j^+i2io 


442  NATURAL  LOGARITHMS. 

with,  the  remainder  Riq  omitted.  It  will  be  useful  to 
observe  how  the  neglect  of  this  remainder,  and  of  the 
errors  in  the  seventh  decimal  figures  of  the  successive 
terms,  might  affect  the  final  result.  Now,  by  the  formula 
fori2„, 

i2io<j^<. 000000005, 

and  the  neglect  of  this  number  could  not  affect  the 
seventh  decimal  figure ;  moreover,  the  aggregate  of  the 
errors  in  the  several  terms  could  not  exceed  8  x  .5,  or  4, 
in  the  seventh  place,  and  this  could  not  affect  the  fifth 
decimal  figure.  Hence  the  above  computation  gives  an 
approximate  value  of  e  correct  to  at  least  five  places  of 
decimals. 

307.   The  following  are  important  examples  of  limits 
that  involve  exponential  and  logarithmic  functions. 

Ex.  1.   Find  the  Hmit  of  (a'  —  l)/x  when  x  =  0, 

/yx  1         fixlna 1 

Since =  ^— •  In  a, 

X  xlna 

...  «»i'^«l^  =  l„a.  ■        [Art.  303.] 

Ex.  2.   Find  the  limit  of  (a*  —  a^)/(x  —  n)  when  a  =  n. 
a*  —  a**      „„  a*-"  —  1 


Since 


x  —  n  X  —  n 


.^Tio^-i^=^-'  CEX.1.] 

Hmit^-^^^„j^^^ 


x  =  n 


X 


Ex.  3.   Find  the  limit  of  a^/x  when  a;  =  oo,  a  >  1. 

Whatever  rational  positive  value  x  may  have  we  can  always 


1, 


NATURAL  LOGARITHMS.  443 

express  it  as  the  sum  of  a  positive  integer  n  and  a  positive  finite 
proper  fraction  /,  and  we  may  therefore  write 

9l.  —  ^^^^  —  of     ^    ^ 
X      /+  n         /+  n  w' 

in  which  /  and  af  remain  positive  and  finite  for  all  positive  values 

of  X.    But,  because/  is  finite, 

limit      n     _   limit        1 

and  therefore 

limit  g^_    f  limit  a^ 

in  which  w,  always  a  positive  integer,  approaches  oo  by  successive 
integral  steps. 

Also,  since  by  hypothesis  a  >  1,  we  may  write 
a  =  l  +  A,   ^>0; 
and  hence,  by  the  binomial  theorem  for  a  positive  integral  index, 

a«  =  1  +  w/i  +  ^^^  ~  ^^  h^  +  nx  (other  positive  terms), 

—  =  i+^4-—  (n-l)+  other  positive  terms, 
n      n  2 

and 

limit  a_"  ^  limit  ^  (^  _  j)  +  other  positive  terms  =  oo. 
n  =  oo^      71  =  002 

.     limit  a^  _  „f  ^  ^ 

a;  =  oo  a; 

Ex.  4.   Find  the  limit  of  y  "log  y  when 
?/  =  +  0,  a>l. 
Let  y  =  a-"  ;  then 

y  «log  y  =  —  xa-*  =  —  x/a*, 
and  a;  =  +  00  when  y  =  +  0.    But 
limit 


f-^)  =  -^=0.  [Ex.3.] 


limit 


••y=+0y''^°S2^=^- 


444  NATURAL  LOGARITHMS. 

Ex.  5.   rind  the  limit  of  x*  when  x=  +  0. 
By  Art.  291,  a;*  =  6=^1°  =»; 

.      limit  limit     .m^.^o-i  pEx  41 

Ex.  6.   Eind  the  limit  of  (x  -  1)1"*  when  a:  =  1. 
We  may  write 

(X  -  1)1"=^  =  {(X  -  l)=-l}(lnx)/(x-l). 

But  ^"^l  (X  -  l)-i  =  1,  [Ex.  4.] 

and  ^^^^J(lnx)/(x-l)=l.  [Art.  301.] 

.-.    ^^^i*^(x-l)ln-=ll  =  l. 


308.  Exponential  Indeterminate  Forms.  In  order  to  dis- 
cover under  what  conditions  the  function  u"  will  assume 
an  indeterminate  form,  as  in  Exs.  5  and  6,  Art.  307,  we 

write 

f^W,  logf=vlogu, 

the  logarithms  being  taken  in  any  system,  and  seek  the 
conditions  that  make  vlogu  indeterminate,  and  these 
will  also  be  the  conditions  that  make  /or  u"  indetermi- 
nate. Since  log  1  =  0,  log  co  =  oo,  and  log  0  =  —  oo,  these 
conditions  obviously  are 

-y  log 2^  =  (  ±  oo)  X  0,  or  w"  =  1  ±*, 

vlogu  =^  0  x{+co),  OT  u"  =  co^, 

vlogu  =  0  x{—oo),  OT  u^  =  0^. 

These  are  known  as  exponential  indeterminate  forms, 
and  the  list  is  evidently  exhaustive  for  forms  that  result 
from  combinations  of  values  of  u  and  v.     But  of  course 


NATURAL  LOGARITHMS.  445 

either  u  ot  v  may  by  itself  assume  one  of  the  algebraic 
forms  0/0,  co/oo,  0  x  co,  oo  —  oo. 

Ex.  1.   The  limit  of  (1  +  x/n)^  when  n  =  co  (x  =  a  finite  quan- 
tity) is 

Ex.  2.    Find  the  limit  of  xV*  when  x  =  oo. 
Let  X  =  l/z,  then  2  =  0  when  x  =  oo, 

and 

xi/«  =  (i/2)-  =  l/2*. 

Ex.  3.  Find  the  limit  of  (In  x)'-^  when  x  =  1. 
We  proceed  as  in  Ex.  6,  Art.  307,  and  obtain 
(lnx)=^-i  =  {(lnx)^°'}(^-i)/i°*, 

i^^\Vlnx)-^  =  li  =  l. 


EXAMPLES  LXXXIII. 

Evaluate  the  limits  of  the  following  functions  for  the  values  ot  x 
indicated : 

1.  In(x3  —  1)  —  ln(x  —  1)  when  x  =  1. 

2.  In(x2  _  1)+  In /  1+  — ^"j  when  x  =  1. 

3.  In  {2  -  VCl  +  X)}  -  In  {3  -  V(3  x)}  when  «  =  3. 

4.  (x2  -  1)  ln(x  -  1)  when  x  =  1. 
6.    (1  +  l/x2)*  when  x  =  co. 

6.  (1  +  l/xy^  when  x  =  oo. 

7.  x^/^'  when  x  =  1. 


446  NATURAL  LOGARITHMS. 

8.  xi/(a=2-i)  when  x  =  l. 

9.  {(x3  -  l)/(x^  +  1)}''-^  when  x  =  l. 

10.  (x)^^  when  x  =  0. 

11.  (a-^  -  1)^  when  x  =  0.  , 

12.  (a=*  -  a-^)^  when  x  =  0. 

13.  Solve  the  equation  x^''+^  —  2x2*'  —  a;  +  2  =  0. 

14.  Solve  the  equation 

a^  .  x^+i  —  X  ■  a^  -  x''  -\-  a''  =  0, 

in  which  a  is  positive  and  greater  than  1.     (Take  account  of  the 
result  of  Ex.  3,  Art.  307.) 


CONVERGENCY  AND  DIVERGENCY.      447 


CHAPTER   XXXI. 

CONVERGENCY   AND   DIVERGENCY   OF   SERIES. 

309.  A  succession  of  quantities,  formed  in  order  ac- 
cording to  some  definite  law,  if  finite  in  number,  consti- 
tute a  finite  series ;  but  if  their  number  exceed  every  finite 
quantity,  however  great,  they  are  said  to  form  an  infinite 
series. 

We  have  already  seen  [Art.  266]  that,  when  r  is 
numerically  less  than  unity,  the  sum  of  the  n  terms  of 
the  geometrical  progression  a  -f  ar  +  ar^  -f  •  •  •  -f  ar^~^  can 
be  made  to  differ  from  a/(l  —  r)  by  an  arbitrarily  small 
quantity,  by  sufficiently  increasing  n.  Many  other  series 
have  this  property,  —  that  the  sum  of  the  first  n  terms 
approaches  a  finite  limit  when  n  is  increased  ad  infinitum. 

310.  Definitions.  When  the  sum  of  a  finite  number  of 
terms  of  an  infinite  series  approaches  a  finite  limit  (or 
zero),  as  the  number  of  terms  is  increased  ad  infinitum, 
the  series  is  said  to  be  convergent,  and  this  limit  is  called 
the  sum,  or  limit  of  the  series. 

If  the  series  have  ±  oo  as  its  limit,  it  is  said  to  be  di- 
vergent; and  if  it  have  neither  a  finite  nor  an  infinite 
limit,  it  is  said  to  be  neutral  or  indeterminate.* 

*  Series  of  this  last  class  are  also  called  divergent  series  in  many 
text-books  of  algebra. 


448  CONVERGENCY   AND  DIVERGENCY. 

Ex.  1.  The  series  ^  +  I  +  I  -h  •■■  ad  inf.  is  convergent,  and  has 
the  limit  1.     [Art.  285,  Ex.2.] 

Ex.  2.  The  series  1  +  2  +  3  +  •••  acZ  inf.  has  +  oo  for  its  limit, 
and  is  therefore  divergent. 

Ex.  3.   The  series   1  —  1  +  1  —  IH ad  inf.  has  no  definite 

limit,  but  is  1  or  0  according  as  the  number  of  terms  is  odd  or  even; 
it  is  therefore  indeterminate. 

311.  It  is  evident  that  an  infinite  series  whose  terms 
are  either  all  positive  or  all  negative  cannot  be  indeter- 
minate, but  must  have  either  a  finite  or  an  infinite  limit. 

If  each  term  of  an  infinite  series  be  finite  (not  zero) 
and  the  terms  are  either  all  positive,  or  all  negative,  the 
series  must  be  divergent.  For,  if  each  term  be  not  less 
than  a  the  sum  of  n  t^rms  will  be  not  less  than  na,  and 
na  can  be  made  larger  than  any  finite  quantity,  by  suffi- 
ciently increasing  n. 

312.  The  successive  terms  of  an  infinite  series  will  be 
denoted  by  Wi,  U2,  ^3?  •••  w„j  •••?  the  sum  of  the  first  n 
terms  by  17^,  and  the  limit  of  the  series,  if  it  be  conver- 
gent, in  which  case  alone  it  has  a  limit,  by  U.     Thus 

C4  =  ^i  +  W2H hw„, 

and  U  =Ui  +  Wg  +  ^^3  +  •••  <xc?  inf. 

We  shall  frequently  omit  the  words  ad  inf.,  and  denote 
the  limit  of  the  entire  series  by  Wi  + 1^2  +  ^3  +  •••• 

313.  In  order  that  the  series  Ui-\-U2-\- f-^n  +  *** 

may  be  convergent,  it  is,  by  definition,  necessary  and 
sufficient  that  each  of  the  finite  sums  Un,  C^+u  C^«x2> 
etc.,  shall  approach  the  common  limit  U,  as  n  increases. 
Hence   U^,    U^+i,   Un+2,  etc.,  must  differ  from   U,   and 


CONVERGENCY  AND  DIVERGENCY.      449 

therefore  from  one  another,  by  quantities  that  converge 
to  zero,  as  n  increases  without  limit. 

But  Un+i-   Un=U^^j 

U-    Un  =  W„+l  +  W„+2  +  Wn+3  +  •••• 

Hence :  In  order  that  a  series  may  he  convergent^  the 
(n  +  1)^^  term,  and  also  the  sum  of  any  number  of  terms 
beginning  with  the  {n-\-l)th^  must  have  zero  as  a  limit 
when  n  is  increased  ad  infinitum. 

For  example,  the  series  1/2  +  1/3  +  1/4  +  •••  +  1/n  +  •••  can- 
not be  convergent,  although  limit  „=oo(l/w)  =  0 ;  for  the  sum  of  n 
terms,  beginning  at  the  (n  +  l)th,  is 

1      +_!_,+      1+...+    1 


n+1     n+2     n+3  2n 

which  is  greater  than  (1/2  n)  x  n,  that  is,  than  1/2. 

314.  Tests  for  Oonvergency.  The  following  are  the  most 
important  criteria  for  the  convergency  of  infinite  series. 
It  is  convenient  to  state  them  as  applying  only  to  series 
whose  terms  are  all  positive  (Criterion  IV.  and  the  cases 
in  which  negative  terms  are  chai-acteristic  of  the  series 
excepted).  But  they  are  also  effective  when  applied  to 
series  whose  terms  are  all  negative,  and  (provided  the 
signs  of  all  negative  terms  be  changed)  to  that  class 
of  series,  containing  both  positive  and  negative  terms, 
which  remain  convergent  when  all  the  negative  terms 
are  made  positive.  For,  if  a  series  be  convergent  when 
all  its  terms  are  positive,  it  is  obviously  convergent  when 
either  all  or  some  of  its  terms  are  negative. 

27 


450      CONVERGENCY  AND  DIVERGENCY. 

For  example,  if  m^  +  m.^  +  Wg  +  ••• 

is  convergent,  so  ig  —  (Wj  +  M2  +  %  +  *•*)» 

and  also  w^  —  Wj  +  %  —  W4  +  •••. 

The  term  series  means  always  infinite  series  unless 
specific  statement  to  the  contrary  is  made. 

315.  Test  by  the  difference  between  the  terms  of  the 
given  series  and  the  corresponding  terms  of  a  convergent 
series. 

I.  A  series  {whose  terms  are  all  positive)  is  convergent 
if,  after  any  particular  term,  each  of  its  terms  is  less  than 
the  corresponding  term  of  a  known  convergent  series. 

Let  there  be  two  series 

XJ=,S-\-u^-\-U2-\-u^-\-'",    V^Vi  +  v.,  +  v.^-\ , 

in  which  S  is  the  sum  of  a  finite  number  of  the  terms  of 
U;  suppose  V  to  be  convergent,  and  that  u^  <  v^  for  all 
values  of  r.     Then  U—S  is  certainly  less  than  F,  that  is, 

U<  F+  S, 

and  since  Fand  8  are  finite,  so  also  is  XJ)  and  CT  must 
have  a  limit,  for  all  its  terms  are  positive. 

II.  It  needs  only  a  slight  modification  of  this  reason- 
ing (which  the  student  can  easily  make)  to  prove  that 
an  infinite  series  {having  only  positive  terms)  is  divergent 
if,  after  any  particular  term,  each  of  its  terms  is  greater 
than  the  corresponding  terms  of  a  divergent  series. 


CONVERGENCY  AND  DIVERGENCY.      451 


Ex.  1.    Apply  the  test  for  convergency  to  the  series 


1      1.2      1.2.3      1.2.3.4 

The  terms  of  this  series  are,  in  the  order  of  occurrence,  less 
than  the  corresponding  terms  of  the  geometrical  progression 

1+ 1 +^j_,+    1     . 


1      1-2      1.2.2      1.2.2.2 

whose  common  ratio  is  ^,  and  which  is  therefore  known  to  be 
convergent.     The  given  series  must  therefore  also  be  convergent. 

Ex.  2.  Test  for  convergency  the  series 

1  +  3  +  ^+33      3*      3^     _ 
2!3!4!6! 

After  the  fifth  term  each  term  of  this  series  is  less  than  the 
corresponding  term  of  the  geometrical  progression 

S6  Rd  37 


4  .  4  1      42 .  4  !      48 . 4  I 

whose  common  ratio  is  3/4,    The  given  series  is  therefore  con- 
vergent. 

316.  Test  by  the  ratio  of  the  terms  of  the  given  series 
to  the  corresponding  terms  of  a  convergent  series. 

III.  If  the  ratio  of  the  corresponding  terms  of  two  series 
(of  positive  terms)  be  always  finite,  the  series  mil  both  be 
convergent  or  both  divergent. 

Let  the  series  be  respectively 

?7=Wi  +  W2  +  %H J    V=v^-{-v^-\-Vz-\-"'. 

Then,  since  all  the  terms  are  supposed  to  be  positive, 
^/Fmust  be  greater  than  the  least  and  less  than  the 
greatest  of  the  fractions  Ui/vi,  U2/V2,  u^/v^,  •••  [Art.  170, 


452  CONVERGENCY   AND   DIVERGENCY. 

Theorem  II.].     Hence   C7/F  is  finite;  and,  therefore,  JJ 
is  finite  or  infinite  according  as  Fis  finite  or  infinite. 

Ex.  1.    The  two  series 

2-3     4-5     6-7  2  7i(2w+l)  ' 

3  +  5  +  7  +  -+2^r^+-' 

are  both  convergent  or  both  divergent.  For,  the  ratio  of  the  n*^ 
terms  is  (n  +  l)/2  w,  or  1/2  +  1/w,  which  cannot  be  greater  than 
3/2  or  less  than  1/2  for  values  of  n  between  1  and  co. 

Ex.  2.  Prove  that  the  two  series  of  Ex.  1  are  divergent  by  com- 
paring the  terms  of  the  second  one  with  the  terms  of  the  divergent 
series  1  +  1/2  +  1/3  + h  l/n  -\ . 

317.   Test  by  ratio  of  convergence. 

IV.  A  series  is  convergent  if,  after  any  particular  term, 
the  absolute  value  of  the  ratio  of  each  term  to  the  preceding 
is  always  less  than  some  fixed  finite  quantity  which  is  itself 
less  than  unity. 

Let  the  terms  be  supposed  to  be  all  positive,  and  in 
accordance  with  the  conditions  here  assigned,  beginning 
with  the  (r  +  1)*^  term,  let 

u^+\/u^  <  k,  w^+a/'^r+i  <  ^?  u^+Ju^^2  <  ^j  etc.,  where  A;  <  1. 

Then  u^^^  <  ku^,  u^j^2  <  ^^^r+i  <  ^^^r?  ^r+s  <  ^^^rj  3,nd  so  on 
ad  infinitum.     Hence 

Ur  +  u,+i  +  Wr+2  +  '••  <  u,{l  +  A)  +  A^^  +  -)        [Art.  279.] 
<  uj{l  -  k),  '.'  k  <  1.     [Art.  266.] 

Therefore,  the  limit  of  the  series  beginning  with  the 
7*^  term  is  finite,  and  since  the  sum  of  any  finite  number 
of  terms  is  finite,  the  entire  series  must  be  convergent. 


CONVERGENCY  AND  DIVEKGENCY.      453 

This  demonstration  remains  valid  for  series  having 
both  positive  and  negative  terms,  if  only  u^  u^^  etc.,  be 
replaced  by  abv  u^,  abv  u^^^,  etc.,  where  abv  is  an  abbre- 
viation for  "absolute  value  of";  for,  if  the  series  be 
convergent  when  all  its  terms  are  positive,  it  is  certainly 
convergent  when  some  of  them  are  negative. 

We  shall  hereafter  frequently  use  the  abbreviation  abv 
to  denote  absolute  value  of. 

Def.  The  ratio  u^+i/u^  is  called  the  ratio  of  convergence 
of  the  series  Ui  +  u^-^-  u^  +  •••, 

V.  A  series  {of  positive  terms)  is  divergent  if  after  any 
particular  term,  the  ratio  of  each  term  to  the  preceding  is 
either  equal  to  or  greater  than  unity. 

If  every  term  after  the  r^  be  equal  to  u^  then 

W,+l  +  Wr+2  -\ h  y^r+n  =  flU^ 

or,  if  after  the  r*^  term  the  ratio  of  convergence  be 
greater  than  1,  then 

W,+1>W„     Wr+2  >  W,+i  >  W^,     "'U,^^>U^ 

that  is, 

W,+l  -f  Wr+2  H h  ^^r+n  >  nU^ 

In  either  case  the  sum  of  these  n  terms  can  be  made 
greater  than  any  iinite  quantity  by  sufficiently  increasing 
71.  Hence  this  part  of  the  series,  and  therefore  the  entire 
series,  is  divergent. 

Ex.  1.   The  ratio  of  convergence  of  the  series 

1+1+1+...+A+... 

2      22     28  2" 

is  {(n  +  l)/2«+i}/(n/2«)  =  (n  +  l)/2  n,  and  this  is  less  than  f  for 
all  values  of  n  greater  than  2.     Hence  the  series  is  convergent. 


454      CONVERGENCY  AND  DIVERGENCY. 

Ex.  2.    The  ratio  of  convergence  of  the  series 
2  22         23         2*  2« 

1.22.33.44.5  w(w+l) 

is  2(n  ^  l)/(w  +  1),  which  is  greater  than  1  for  all  values  of  n 
greater  than  2.     Hence  the  series  is  divergent. 

Ex.  3.  The  ratio  of  convergence  of  the  series 
12  _|.  22x  +  32x2  +  ...  +n2x«-i  +  — 
is  x(n  +  l)2/n2,  that  is,  (1  +  1/nyx. 

Now  if  X  be  positive  and  less  than  1,  and  any  fixed  quantity 
k  be  chosen  between  x  and  1,  the  ratio  of  convergence  will  be  less 
than  k  for  all  terms  after  the  first  one  that  renders 

{l  +  l/nyx<k, 
that  is,  for  all  terms  after  the  n}^,  where 

n>  y/x/{^k-  y/x), 
the  square  roots  being  taken  positively.     Hence  the  series  is  con- 
vergent if  ab  V  X  <  1 . 

If  X  =  1  the  series  is  I2  +  22  +  32  +  ..-,  which  is  obviously  diver- 
gent, and  it  is  therefore  also  divergent  if  x  >  1. 

When  an  infinite  series  is  such  that,  after  a  finite 
number  of  terms,  its  ratio  of  convergence  u^^i/u^  is 
always  less  than  unity,  but  approaches  unity  as  a  limit, 
the  test  by  ratio  of  convergence  fails.  Further  special 
investigation  is  then  necessary. 

Ex.  4.   The  ratios  of  convergence  of  the  series 

i  +  i  +  i  +  i  +  ...  +  L+    1     ■ 


2      3      4  n      n+1 

1  +  1  +  1+1+.. .  +1  + }: +  ... 

22      3-'      42  n2      (w  +  1)2  ^ 

are  n/(n  +  1)  and  ri- /  {11  +  1)2  respectively,  and  each  of  them  is 
less  than  unity  for  all  finite  values  of  w,  but  approaches  unity  as  a 
limit  when  n  is  increased  without  limit.  But,  as  will  be  immedi- 
ately proved  in  the  next  article,  the  first  series  is  divergent,  the 
second  convergent. 


CONYERGENCY   AND   DIVERGENCY.  455 

318.  The  convergency  or  divergency  of  many  series 
may  be  determined  by  the  methods  of  comparison, 
explained  in  Arts.  315,  316,  with  the  aid  of  the  series 
1/1*  +  1/2*  + 1/3*  H-  .-.  as  a  standard.  For  this  purpose 
it  is  important  to  know  for  what  values  of  k  this  standard 
series  is  convergent. 

VI.  The  series  1/1*  + 1/2*  + 1/3*  +  •••  is  convergent 
when  k  is  greater  than  unity,  and  is  divergent  when  k  is 
equal  to  or  less  than  urdty. 

(i.)  If  k>l,  since  each  term  is  then  less  than  the 
preceding, 

i  +  i<? 

2*     3*     2*' 

4*     5*     6*     7*     4*' 

1.  1  .  i  .    1    .  !_  ,    1    .  _!  .  A^-.:? 
8*      9*      10*     11*      12*     13*  "^  14*      15*     8*" 

and  in  general 

i+_i_  + 1 <?:. 

2nA^(2'*  +  l)*  (2"+^-l)*       2«* 

Hence,  the  entire  series  is  less  than 

12       4       8  2** 

—  +  —  4-  —  -I-  —  -4-  •••  4-—-+  ••• 

1*  ^  2*     4*      8*  2"*  ' 

that  is,  it  is  less  than 

1+^+^-+-^  +  -+^-+... 

which  is  a  geometrical  progression  whose  common  ratio, 
l/2*~\  is  less  than  unity,  since  A;  >  1.  The  given  series 
is  therefore  converijent. 


456      CONVERGENCY  AND  DIVERGENCY. 

(ii.)    If  A;  =  1,  the  series  may  be  written  in  the  form 

l  +  i  +  (i  +  i)  +  (i  +  i  +  l  +  i)+- 

V2«-i  +  1      2"-^  +  2  ^  2^y 

with  its  terms  so  distributed  into  groups  that  each  group 
(within  brackets)  is  greater  than  i  Hence  2"  terms  of 
the  given  series  is  greater  than  n-\-l  terms  of  the  series 
^  +  2"+i  +  i+"*j  *^^^^  ^s>  greater  than  1  +  i n,  which 
may  be  made  larger  than  any  finite  quantity  by  suflB.- 
ciently  increasing  n.     The  series 

is  therefore  divergent. 

(iii.)  If  A;<  1,  then  each  term  of  l/P+l/2*+l/3*+--- 
is  greater  than  the  corresponding  term  of  the  divergent 
series  1  +  i  +  ^  +  •••.  The  original  series  is  therefore 
also  divergent  in  this  case.  This  completes  the  proof  of 
our  theorem. 

Ex.  1.   The  series 

is  divergent. 

For  since  2w/  (w^  +  1)  >  1  /w  for  all  values  of  n  greater  than  1, 
.-.  S2n/Cn2  +  l)>Sl/w, 
and  S  1/n  is  divergent. 

Ex.  2.  The  series  2  (w  +  2)/(n^  +  1)  is  convergent. 
For,     (w  +  2)/(w3  +  i)<  (n  +  2)/n^<3n/n^  =  Z/n\ 
.-.  S  (w  +  2)/(n3  +  1)<3  S  l/w2, 
and  S  l/n^  has  been  shown  to  be  convergent. 


CONVERGENCY  AND  DIVERGENCY.      457 

319.  Def.  If  a  series  having  both  positive  and  nega- 
tive terms  remain  convergent  when  all  its  negative  terms 
are  made  positive,  it  is  said  to  be  absolutely  convergent. 

The  criteria  of  Arts.  315-318  are  effective  only  when 
applied  to  absolutely  convergent  series.  And  there  is  a 
class  of  series,  designated  as  semi-convergent,  whose  con- 
vergency  is  due  to  the  presence  of  negative  terms,  and 
which  become  divergent  when  all  the  terms  are  taken 
positively.  Other  tests,  not  yet  considered,  are  then 
necessary. 

For  example,  as  will  be  shown  in  the  next  article,  the  series 
1  -  ^  +  i  -  i  +  •••  is  convergent,  although  1  +^  +^  +^  +  ...  is 
divergent. 

320.  Test  for  semi-convergency. 

VII.  A  series  whose  terms  are  alternately  positive  and 
negative  is  convergent  if  each  term  is  less  than  the  preceding 
and  the  n^^  term  has  zero  as  a  limit  when  n  increases  ad 
infinitum. 

The  general  series  conforming  to  these  conditions  is 

U=Ui  —  1*2  +  ^3—^4  H ±^n  T  ^Wl  ±  •••> 

with        Wi  >  Wg  >  "^  >  "^4  >  •••  >  ^n  >  Wn+1  >  •••  >  0, 

J  limit  r. 

and  „  =  ^«n  =  0; 

and  this  we  may  write  in  either  of  the  forms 

U^—  {lh—  Us)  —  (^4  —  y's) ) 

from  which  it  is  at  once  evident  that  the  limit  of  the 
series  is  greater  than  w,  —  Wj  and  less  than  u^,  and 
is  therefore  finite.  It  is  also  similarly  evident  that 
abv  (U  —  Un)    is  intermediate    to  abv  (w„+i  —  Wn+2)   a-iid 


458      CONVERGENCY  AND  DIVERGENCY. 

SihYUn+i  (abv  =  absolute  value  of),  and  tliat  therefore, 
since  limit  u^  =  0,  .  ^ 

limit  .:rx      Txs      ^  V-aT/V^ 


The  series  must  consequently  be  convergent.        /^ 


„r;(^-^")=o- 


Ex.  1.  The  series  1  -  1/2  +  1/3  -  1/4  +.••  satisfies  the  condi- 
tions of  this  test,  and  is  therefore  convergent. 

Ex.  2.  The  series  2/1  -  3/2  +  4/3  -  5/4  +  ...  cannot  be  less 
than  1/2  nor  greater  than  2,  but  it  is  not  convergent,  for  the  n^^ 
term,  which  is  (?i  +  l)/w,  has  not  0,  but  1,  as  its  limit  for  w  =  oo. 

321.  Power  Series.  The  infinite  series  whose  general 
term  is  a^x""  {a^  independent  of  x)  is  called  a  power  series 
in  X.  By  Criterion  IV.  [Art.  317]  this  series  is  absolutely 
convergent  if  abv  xa^+i/a^  remain  less  than  1  in  the 
limit  when  n  =  cc,  that  is,  if  x  satisfy  the  condition 
abv  cc  <  abv  limit  (a^/a^+i)  when  n  =  oo.     Hence  : 

YIII.    The  power  series 

ao  +  aiX  +  aacc^  H f-  a^af  H 

IS  absolyitely  convergent  for  all  values  of  x  that  satisfy  the 
condition 

abva!<aby^™^_^(a„/a„+0- 

322.  If  a  power  series  be  only  semi-convergent,  since 
the  successive  terms  of  every  convergent  series  must  ulti- 
mately decrease  in  absolute  value  [Art.  313],  the  ratio 
abv(iKa,i+i/a„)  must  ultimately  still  remain  less  than  1  for 
finite  values  of  n.  But  if  abv(£ca„+i/a„)  remain  less  than 
1  in  the  limit  when  n  =  ao,  the  series  is  absolutely  con- 
vergent [Criterion  VIII.].     Hence,  semi-convergence  can 


CONVERGENCY  AND  DIVERGENCY.      459 

only  arise  by  the  ratio  of  convergence  being,  in  absolute 
value,  ultimately  less  than  1  for  finite  values  of  n,  but 
having  the  limit  1  when  n  =  go.*     Therefore  : 

IX.    If  the  power  series 

tto  +  «ii»  +  ^2^  +  •  •  •  +  o^n^;**  4- •  •  • 
be  convergent  for  a  given  value  of  x,  it  is  absolutely  conver- 
gent for  every  numerically  smaller  value  of  x;  and  if  it  be 
only  semi-convergent  for  the  given  value,  it  is  divergent  for 
every  numerically  larger  value. 

Upon  the  following  corollary  rests  the  proof  of  the 
principle  of  indeterminate  coefficients  which  will  be  dis- 
cussed in  the  next  chapter. 

Cor.  If  the  series  CTq +  %^  +  ot2^+ •••  be  convergent 
for  any  finite  value  ofx, 

Proof:  From  the  identity 
tto-f  ttiJcH-  a^x^  -\-  •••  =  ao-\-x(ai  -\-a^-^a^-\-  •••) 
it  is  evident  that  for  finite  values  of  x  that  render 
aQ  +  aiX-\- '"  convergent,  the  series  a^ -\- a.jX -\- • --  is  also 
convergent ;  hence,  by  Criterion  IX.,  aj  -h  ag^  +  •••  is  con- 
vergent, that  is,  finite  when  a;  =  0.  Therefore,  when 
a;=0, 

aiX  -f  ajX^  4-  ...  =  0  X  {finite  quantity)  =  0. 

323.  The  Binomial,  Exponential,  and  Logarithmic  Series  are 
the  three  infinite  series  of  greatest  importance  in  mathe- 
matical analysis.     We  may  now  test  their  convergency. 

*  Harnack,  Differential  and  Integral  Calculus,  Cathcart's  trans- 
lation, §  44,  page  73. 


460      CONVERGENCY  AND  DIVERGENCY. 


Ex.  1.  In  the  binomial  series, 

mCm  —  l)                  m(m —  l)---('m  —  n  +  1) 
l  +  mx+        2!     ^  +  •••  +  ^ ^^T! ^^"  +  ••*' 

the  number  of  terms  is  finite  when  m  is  a  positive  integer ;  but 
when  m  is  not  a  positive  integer,  no  one  of  the  factors  m,  m  —  1, 
?/i  —  2,  •••  can  be  zero,  and  the  series  will  not   terminate.      It  is 
with  this  second  alternative  that  we  are  at  present  concerned. 
The  ratio  of  convergence  is 

Un+i     m  —  n  +  1 
Un   ~  n 


and  its  absolute  value,  as  n  is  increased,  becomes  more  and  more 
nearly  equal  to  x.  If,  therefore,  abvcc<l,  then,  either  from  the 
beginning  or  after  a  finite  number  of  terms,  abv(Wn+]/Wn)  will  be 
and  remain  less  than  1,  whether  the  sign  of  Un-\-\/Un  be  positive  or 
negative,  and,  by  Criterion  IV.  [Art.  317],  the  series  formed  by 
adding  together  the  absolute  values  of  the  terms  will  be  conver- 
gent. Hence,  for  all  values  of  x  numerically  less  than  unity,  the 
binomial  series  is  absolutely  convergent. 

If  abv  X  be  greater  than  unity,  abv  (Un+i/un)  will  ultimately 
become  and  remain  greater  than  1  and,  by  Criterion  V. ,  the  series 
will  be  divergent. 

[It  may  be  shown  that  this  series  is  convergent  when  x  =  1, 
provided  w  <  —  1,  and  when  x  =  —l,  provided  m  >  0.  See 
Treatise  on  Algebra,  Art.  338.] 

Ex.  2.  The  exponential  series  is 

Its  ratio  of  convergence  is  x/n,  and  this  is  numerically  less  than 
unity  for  all  terms  after  the  first  one  that  renders  x  numerically 
less  than  n.  This  series  is  therefore  absolutely  convergent  for  all 
finite  values  of  x. 

Ex.  3.   The  logarithmic  series  is 


3       4  n 


CONVERGENCY   AND  DIVERGENCY.  461 

and  its  ratio  of  convergence  is 

\       n+l) 


Un+i  _        xn 
tin  n+1 


This  is  numerically  less  than  unity  for  all  values  of  x  that  are 
numerically  less  than  unity,  and  for  such  values  the  series  is  there- 
fore convergent. 

If  oj  =  1,  the  series  becomes  1  —  ^4-^  —  ;^+---,  which  is  con- 
vergent by  Criterion  VII. 

If  X  =  —  1,  the  series  becomes  -(l  +  |  +  i  +  i+  —)»^liich  was 
shown  to  be  divergent  in  Art.  318. 

Thus,  the  logarithmic  series  is  absolutely  convergent  when  x  has 
any  value  between  —  1  and  +  1  and  semi-convergent  when  x  =  1, 
but  divergent  in  all  other  cases. 

324.  Product  of  Series.  If  finite  portions  of  the  two 
series 

V=Vi-\-V2  +  Vs+  •••  +v„-f  ... 

be  multiplied  together  by  the  algebraic  rule  for  the  mul- 
tiplication of  two  multinomials,  a  result  will  be  obtained 
which  may  be  arranged  in  the  form 

P  =  UiVi  +  (^1^2  +  U2V1)  +  (^^i-^s  +  WjjVa  -f  U3V1)  -f-  ••• 
4-  {uiV^  +  U2V^_-i  -^ -f  u„_iV2  +  u^Vi)  +  .... 

Whenever  these  three  series  are  simultaneously  conver- 
gent, the  third  is  defined  as  the  product  of  the  other  two. 

For  our  present  purpose  we  assume  that  the  first  and 
second  series  are  absolutely  convergent,  denote  their 
limits  by  U  and  V  respectively,  and  attach  the  usual 
interpretations  to  U„,  F„,  etc.  Let  the  third  series  be 
denoted  by  P.  It  is  required  to  prove  the  following  very 
important  theorem : 


462  C0NVERGEI5CY   AND   DIVERGENCY. 

If  U  and  V  be  absolutely  convergent,  then  P  will  be  a 
convergent  series  whose  limit  is  equal  to  the  limit  of  the 
product  Ux  V. 

(i.)  If  all  tlie  terms  in  CT'and  Fare  positive,  then  the 
product  C/^n  X  V2n  will  contain  all  the  terms  of  p2n  pl^s 

others,     such    as     U2V2n,    ^3'y2n-l7    ^3^2n}    '^2n'^2nJ    ^^C.,    aud    P2n 

will  contain  all  the  terms  of  the  product  Un  X  F„  plus 
others,  such  as  UiV2n,  ^2^2n-i)  ^3'^2n-2)  etc."*     Hence 

U2nX    V2n>P2n>    ^4  X    F„ , 

and  this  inequality  remains  true,  however  great  a  finite 
value  n  may  have.  But  in  the  limit,  when  n  =  oo, 
JJ^n  =  Un=  U,  and  F2„  =  F„  =  F  Hence,  when  all  the 
terms  are  positive, 

Ux  V=P. 

(ii.)  If  U  and  F  contain  both  positive  and  negative 
terms,  change  the  signs  of  all  negative  terms  and  denote 
the  resulting  series  by  C7'  and  F;  and  let  P'  be  the 
series  formed  from  U  and  F'  in  the  same  way  as  P  is 
formed  from  C/'and  F 

Then  U2n  X  F2n  —  An  cannot  be,  in  absolute  value, 
greater  than  U\n  X  V\n  —  P^'iny  for  the  series  which  the 
two  expressions  stand  for  differ  only  in  the  respect  that 
the  former  contains  some  negative  terms  that  are  posi' 
tive  in  the  latter.     But,  by  the  proof  in  (i.), 

*  The  student  can  make  this  quite  evident  by  writing  out  all  the 
terms  of  these  products  for  small  values  of  n. 


CONVERGENCY   AND  DIVERGENCY.  463 

hence  it  follows  again  that 

Ux V=P* 
This  completes  the  proof  of  the  theorem. 

EXAMPLES  liXXXIV. 
Apply  the  proper  test  of  convergency  to  each  of  the  following 


series : 


1. 22. 33. 4  w(n+l) 

1,1.1..  1 


X      ^ 


x-\-\     x-f3     aj  +  5  2c  +  2n 

•    1*    -3*     6*  (2n-l)* 

6     1-2     2.3     3-4     4-5 
*3.4     4. 56. 6     6.7 

«       1 • 2    ,    2.3        34  ^  4.5  ^       ' 


32.42     42.52      52.52     62.72 

7     a;-l      y.  —  2      x  —  3  ,^  —  ^  ■ 

x+1      x+2     x+3     x+4 

The  following  expressions  denote,  in  each  case,  the  n*^  term  of 
an  infinite  series.  By  applying  the  proper  tests,  determine  whether 
the  series  thus  indicated  are  convergent  or  divergent. 


8.    I 9.    ?*(!L±il. 


10. 


1 


/o.     .  u.     — » ^«  xu.     • 

2?i(2^i-l)  2"  l  +  n2 

*  This  result  is  also  true  when  only  one  of  the  factor  series 
Z7,  V,  is  absolutely  convergent,  the  other  being  semi-convergent. 
If  neither  U  nor  V  be  absolutely  convergent,  nothing  is  certain 
about  the  convergency  of  P.  [See  Chrystal,  Algebra,  Vol.  II., 
Chapter  XXVI.,  §  14.] 


464  CONVERGENCY  AND  DIVERGENCY. 


l  +  n2 

an2  +  6 

,     1  +  n 

16. 

;.  ^  +  1. 

17. 

X" 

n2 

l  +  n2 

L         1 

18. 

xn-l 

an  +  b 

n2 

19.        ^" 


20. 


l  +  x«     ' 
21    (^  +  1)^^ 

22.  -^:^. 


INDETEKMINATE  COEFFICIENTS.  465 


CHAPTER   XXXII. 

Indeterminate  Coefficients. 

325.  In  Art.  148  it  was  proved  that  if  a  rational  inte- 
gral function  of  x  vanish  when  x=  a,  it  is  divisible  by 
X  —  a  without  remainder.     Let  the  function  be 

/  {x)  =  ax"  +  baf"-^  +  caf'^  -\ [-1, 

and  let  the  division  by  a;  —  a  be  performed.  It  is  clear 
that  the  first  terra  of  the  quotient,  namely,  the  term  of 
highest  degree  in  x,  will  be  ax""'^ ;  and  since  dividend  = 
divisor  x  quotient, 

.-.  /(a;)  =  (a;-«)(aa;«-i-f-..). 

Suppose  that/(«)  also  vanishes  whena;=/?,  [/8  not  =a], 
then  the  product  (a;  —  a  )  x  (aaj**"^ -f  •  •  • ) ,  and  therefore 
either  (x  —a),  or  (ax""'^  +  •••),  must  vanish  when  x  =  p. 
But  )8  —  a  is  not  zero ;  hence  {ax'*''^  +  •••)  vanishes  when 
Xz=  f3,  and  is  therefore  divisible  by  x  —  ft-,  and  if  the 
division  be  performed,  the  first  term  of  the  quotient  will 
be  ax'''^ 

.-.  f{x)  =  (x-a){x-  13)  {ax'*-'  +...)• 

In  general,  if  there  be  r  values  a,  /?,  y,  8,  •••  of  a;  for 
which  /  (a;)  vanishes,  r  repetitions  of  this  process  will 
obviously  produce 

f{x)  =  {x-a)(x-(3){x-y)(x-8)"'{ax'*-^-\--.). 

Finally,  if  ax""  -f-  bx^~^  +  •••  vanish  for  n  different  values 
2a 


466  INDETERMINATE   COEFFICIENTS. 

of  X,  it  has  n  factors  x  —  a,  x  —  p,  etc.,  and  a  final  factor 
ax^-'',  or  a.     Therefore,  under  this  hypothesis, 
f{x)  =  a{x-  a)  {x  _  ^)  (x  -  y) ..., 
in  which  there  are  n  binomial  factors. 

Two  or  more  of  these  binomial  factors  may  of  course 
be  identical.  Should  there  be  a  factor  of  the  r*^  degree, 
say  (x  —  ay,  it  is  evident  that  the  number  of  remaining 
factors  will  he  n  —r. 

326.  A  ratioyial  integral  function  of  the  n*^  degree  in  x 
cannot  vanish  for  more  than  n  values  of  x,  unless  the  co- 
efficients of  all  the  powers  of  x  are  zero. 

For,  if/ (a;),  being  of  the  form 

ax""  +  hx""-^  +  cx''-'^  H , 

vanish  for  the  n  values  a,  /5,  y ...,  it  must  be  equivalent  to 
a{x  —  a){x  —  li){x  —  y)'-'. 

If  now  we  substitute  for  x  any  other  value,  k  suppose, 
different  from  each  of  the  n  values  a,  /3,  y,  etc. ;  then 
since  no  one  of  the  factors  k  —  a,7c  —  /S,  etc.,  is  zero,  their 
continued  product  cannot  be  zero,  and  therefore  f{x) 
cannot  vanish  for  the  value  x  =  k,  except  a  itself  is  zero. 

But  if  a  be  zero,  f{x)  reduces  to  bx^~^  +  ccc""^  +  •••?  and 
is  of  the  (n  —  1)*^  degree ;  and  hence  it  can  only  vanish 
for  n  —  1  values  of  x,  except  b  is  zero.     And  so  on. 

When  all  tlie  coefficients  are  zero,  the  function  will 
clearly  vanish  for  any  value  whatever  of  x. 

327.  The  values  of  x  for  which  the  expression 

is  equal  to  zero,  are  the  roots  of  the  equation 

ax""  +  hx""-^  +  ca;"-^  -\ =  0.        [Art.  91.] 


rNDETERMINATE  COEFFICIENTS.  467 

Hence,  by  Art.  326,  an  equation  of  the  n'*  degree  cannot 
have  more  than  n  roots,  except  the  coefficients  of  all  the 
different  powers  of  the  unknown  quantity  are  zero,  in 
which  case  any  value  of  x  satisfies  the  equation. 

328.  If  the  two  expressions  of  the  ?i*^  degree 

aaf  +  baf"-^  +  cx""'^  -\ , 

and  psf  +  qx""-^  +  rx"*-^  H , 

be  equal  to  one  another  for  more  than  n  values  of  x,  it 
follows  that  the  equation 

ax"*  +  6ic"-^  +  c.'c"-^  -f =  px""  +  gic"-^  +  rx""'-  -\ ; 

that  is,  the  equation 

(a—p)x*'-\-  {b  —  q)x''-'-\-  (c  -  r)x"-2  +  ...  =0 
has  more  than  n  roots. 

Hence,  by  Art.  827,  the  coefficients  of  all  the  different 
powers  of  x  must  be  equal  to  zero. 

Thus  «  =i>j  ^  =  ?)  c  =  r,  etc. 

Hence,  if  tivo  rational  integral  functions  of  the  n'*  degree 
in  X  be  equal  to  one  another  for  more  than  n  values  of  x, 
the  coefficient  of  any  poiver  of  x  in  one  function  is  equal  to 
the  coefficient  of  the  same  x>ower  of  x  in  the  other. 

This  is  the  principle  of  indeterminate  coefficients  as 
applied  to  rational  integral  functions. 

329.  When  any  two  rational  integral  expressions, 
which  have  a  limited  number  of  terms,  are  equal  to  one 
another  for  all  values  of  the  letters  involved,  the  condi- 
tion of  the  last  article  is  clearly  satisfied,  for  the  num- 
ber of  values  must  be  greater  than  the  index  of  the 
highest  power  of  any  contained  letter. 


468  INDETERMINATE   COEFFICIENTS. 

Hence  when  any  two  rational  integral  expressions,  which 
have  a  limited  number  of  terms,  are  equal  to  one  another 
for  all  values  of  the  letters  involved  in  them,  we  may  equate 
the  coefficients  of  the  different  powers  of  any  letter. 

330.    If  the  two  infinite  series 

ao -{- a^x  +  a^ -\ \- a^x"" -\ , 

&o  +  b,x  +  b,x'  +^..  +  M"  +  •••, 

be  equal  to  one  another  for  all  finite  values  of  x  for  which 
they  are  convergent,  then 

tto  =  b^,  Oi  =  bi,  ",  a„  =  6„,  etc. 

If  there  be  a  finite  value  of  x  for  which  the  two  series 
are  convergent,  they  are  also  convergent  for  a;  =  0  [Cri- 
terion IX.,  Art.  322],  and  when  a;  =  0,  by  the  corollary 
of  Art.  322, 

aiX  +  acpi?  H =  0, 

and 

biX  +  62^^  +  •  •  •  =  0 ; 

.-.  aQ  —  bQ. 

We  now  have,  for  the  finite  values  of  x  that  make  the 
original  series  convergent, 

a^x  -j-  a^x?  H =  b^x-{-  b^y?  -\ , 

that  is, 

«!  +  acpc  +  a^a^  4-  • . .  ==  6^  -|-  ^2^^  +  b<p(?  -\ . 

But  these  two  series  are  convergent  for  all  the  values 
of  X  that  make  a^ ■\- a^x -\-  -- -  and  bx-\-b^x-\-  -"  conver- 
gent [Art.  322],  and  hence,  by  again  putting  a;  =  0,  we 
obtain 

ai  =  6i. 


INDETERMINATE  COEFFICIENTS.  469 

By  Recessive  repetitions  of  this  process  we  find  ag  =  62? 
ttg  =  63,  and  in  general,  «„  =  6„. 

TMs  is  the  principle  of  indeterminate  coefficients  as  ap- 
plied to  infinite  series. 

APPLICATION  TO   INTEGRAL  FUNCTIONS. 

331.  The  principle  of  indeterminate  coefficients  was 
applied,  without  formal  notice  of  the  fact,  to  Examples  3 
and  4  of  Art.  153.  The  following  are  other  examples  of 
this  class : 

Ex.  1.   Determine  the  values  of  c  and  d  that  will  make 
x^  +  hx'^  +  ex  +  d 
a  perfect  cube  for  all  values  of  x  and  b. 
Assume  x^  +  bx^  +  ex  +  d=(x  +  ky. 
Then,  since 

(x  +  ^•)3  =  x8  +  3  A:x2  +  3  k'^x  +  k^, 
we  must  have 

b  =  3k,  c  =  3k^  d  =  k^,  [Art.  328.] 

whence 

k  =  i,b,  c  =  lb\  d=^b^ 

Ex.  2.  Determine  under  what  conditions  ax'  +  6x*  +  ex  4-  d  is 
divisible  by  px^  +  qx  -{■  r  without  remainder. 

If  the  division  be  performed,  the  quotient  will  consist  of  two 
terms;  namely,  (a/p)x,  and  a  second  term  independent  of  x. 
Hence,  the  quotient  is  of  the  form  (a/p)x  +  A;,  where  A;  is  a  quan- 
tity to  be  determined  by  the  condition  that  the  division  shall  be 
exact.     We  have,  therefore, 

0x8  +  bx^  +  ex  +  d  =  (-x  +  k\  0x2  ^qx  +  r) 


+  (-  +  «A;^«  +  r*, 


470  INDETERMINATE   COEFFICIENTS. 

and  since  this  is  to  be  true  for  all  values  of  a;,  the  coefficients  of 
like  powers  of  cc  in  the  two  members  of  this  equation  must  be  equal 
[Art.  328]  ;  whence 

aq  ar 

b=—-\-pk,  c  =  ---\-qk,  d  =  rk. 

Replacing  k  in  the  first  two  of  these  equations  by  its  value  d/r 
derived  from  the  third,  we  have 

bp  =  aq  +  dj)'^  /r,    cp  =  ar  +  dpq  /r. 

These  are  the  conditions  necessary  and  sufficient  in  order  that 
the  division  may  be  exact. 

Ex.  3.   Transform  x^  +  pxy  +  qy^  into  the  "sum  of  two  squares. 
Assume  u  =(x  +  ■\/qy)h,  v  =(x  —  ■yjqy)k,  and  determine  h  and  k 
by  the  condition  that 

t|2  _j_  1,2  =  3.2  _|_  pxy  -I-  qy^. 

For  this  purpose  we  make  the  coefficients  of  xy  and  y'^  in 

that  is,  in 

(x2  +  gy2)  {h^  +  A:2)  +  2  y/qxyi^^  -  k^)     ,^^ 

equal  to  p  and  q  respectively.     We  thus  obtain 

;i2  +  A:2  =  1^    2  V^C/i^  -  A:2)  =  p, 
or 

yJ'lW        2y/ql'  V2\V        2y/q) 

Hence,  h  and  k  having  these  values, 

x2  +  pxy  +  qy'^  =  {{x  +  ^qy)h}^  +  {(x  -  y/qy)kf. 

Ex.  4.    Find  the  factors  of 

(62c2  _  a*)  (6  -  c)  +  (c2a2  -  M)(c  -  a)  +  (a262  _  c*)  («-&). 

This  expression  vanishes   if  c  =  5,  or  if  a  =  c,  or  if  6  =  a,  and 
therefore  h  —  c,  c  —  a^  and  a  —  h  are  factors.     [Art.  148.  J 

Also,  since  the  expression  is  symmetrical  and  is  of  five  dimen- 


INDETERMINATE   COEFFICIENTS.  471 

sions,  there  must  be  a  fourth  symmetrical  factor  of  two  dimensions, 
which  must  therefore  have  the  form 

Z(a2+  62  +  c2)+  M{bc  +  ca  +  ah), 

The  given  expression  is  therefore  identically  equal  to 
(6  -c)ic-  a) (a  -  h){L{a^  +  6^  +  c^)  +  itf  (6c  +  ca  +  ah)}, 
in  which  L  and  M  are  to  be  determined  by  equating  coefficients  in 
accordance  with  the  principle  of  indeterminate  coefficients. 

The  coefficients  of  a*  in  the  two  expressions  are  respectively 
-  (6  -  c)  and  -  L(b  -c),  and  the  coefficients  of  #  are  b^  -  c^ 
and  &2  _  c2  _  jf  (ft-2  _  c2)  .  hence  L  =  l  and  M=0,  and  the  re- 
quired factors  are  (6  —  c),  {c  —  a),  (a  —  b),  (a^  +  b'^  +  c^). 

EXAMPLES   LXXXV. 

1.  Determine  the  value  of  k  that  will  make  a^  —  «*  +  2x3  +  A; 
divisible  by  ic*  +  x  +  1  without  remainder. 

2.  Determine  the  values  of  p,  q,  and  r  that  will  make  sfi  — 
bpy?  +  bqx^  —  r  divisible  by  (a;  —  c)^  without  remainder. 

3.  If  ax^  +  6x  +  c  and  a'x'^  -\-  6'x  +  c'  have  a  common  factor  of 
the  first  degree,  this  factor,  and  also  the  remaining  (second)  factor, 
in  each  case,  will  be  rational  in  all  the  letters.  Prove  this  by  de- 
termining the  three  distinct  factors.     [Compare  Ex.  3,  Art.  173.] 

4.  Prove  that  if 

ax2  +  2  hxy  +by'^-\-'2,  gx  +  2/y  +  c 
be  expressible  as  the  square  of  a  rational  integral  function  of  x  and 
y  of  the  first  degree,  then 

af  =  gh,  bg  =  hf,  and  ch  —  fg. 
6.  Determine  the  relation  between  a  and  b  that  will  make 
(x  +  ay)"  and  x^  +  bxy  -\-y^  have   a  common   factor  of  the  first 
degree  in  x  and  y. 

6.  Determine  k  such  that  x^  —  y^  —  x  —  Sy  +  k  may  be  the 
product  of  two  rational  factors  of  the  first  degree  in  x  and  y. 

7.  What  value  of  k  will  make  the  three  equations  2x  —  y  + 
3=0,  x  +  y— 1  =  0,  and  x  +  ky  +  I  =  0  simultaneous  in  x  and  y  ? 


472  INDETERMINATE  COEFFICIENTS. 

8.  Supposing  a  and  h  to  be  given  numbers,  determine  a  bino- 
mial expression  (containing  an  arbitrary  factor)  which,  when  sub- 
stituted for  aj,  will  make  ax  +  6^  a  perfect  square  (for  all  values  of 
the  arbitrary  factor). 

9.  Prove  that  if  ax^  +  ho(fi  +  ex  +  d  is  divisible  by  x^  —  A;^,  then 
ad  =  he. 

10.  Find  the  factors  of 

4  a6c (a  +  6  +  c)  4-  he  (h^  +  c^)  +  ca(c2  +  a2)  +  ^^((^2  _}.  52). 

11.  Find  the  factors  of 

&2c2(&  -  c)  +  c'^a^ic  -  a)  +  a262  (a  -  6). 

12.  Find  the  factors  of 

a^(h  -c)+  6*(c-a)+c*(a-6). 

13.  Find  the  cube  root  of 

1  +  3x  +  6x2  +  7  a;3  +  6  X*  +  3x5  +  x6. 

APPLICATION   TO   PARTIAL   FRACTIONS. 

332.  In  Art.  165,  the  process  of  obtaining  a  single 
fraction  as  the  result  of  adding  together  any  number  of 
given  fractions  was  explained.  It  is  sometimes  important 
to  be  able  to  perform  the  converse  operation  of  decom- 
posing a  complex  fraction  into  the  sum  of  a  series  of 
simpler  partial  fractions,  and  for  this  purpose  the  principle 
of  indeterminate  coefficients  is  employed. 

Ex.  1.   Resolve   ^^~  ^ into 

(x-l)(x-2) 
partial  fractions. 

The  sum  of  the  two  fractions  A/(x  —  1),  B/{x  —  2)  will  pro- 
duce a  fraction  whose  denominator  is  (x  —  1)  (x  —  2).  Let  it  there- 
fore be  proposed  to  determine  A  and  B  such  that 

2x-5         ^     A  B     ^  ^(x-2)  +  g(x-l) 

(X- l)(x-2)  ~x-l      x-2~       (x-l)(x-2) 


INDETERMINATE  COEFFICIENTS.  473 

For  this  purpose  it  is  obviously  sufficient  that 

2x-5=(A  +  B)x-i2A  +  B). 

Since  the  left  member  of  this  identity  contains  no  power  of  z 
higher  than  the  first,  both  A  and  B  are  assumed  to  be  constants, 
and  therefore,  by  the  principle  of  indeterminate  coefficients, 

A+  B  =  2,  and  2 ^  +  J?  =  5, 

that  is,  ^  =  3,  and  ^  =  -  1. 

.  2x-5         ^     3      _  ^ 

(x-l)(x-2)      x-1      x-2 

Ex.  2.     Resolve  ^-^t into  partial  fractions. 

x(x-l){x-2) 

Assume  ^l+i =  d+_^  +  _^     and  reduce  the 

x(x-l)(x-2)      X      x-1      x-2 

partial  fractions  to  a  common  denominator.     Then,  the  denomi- 
nators being  omitted,  we  have 

x^  +  l=A(x-  l)(x  -  2)  +  Bx(x  -  2)  -f  Cx{x  -  1). 

We  might  now  expand  and  equate  coefficients,  but  it  is  simpler 
to  proceed  as  follows  : 

Since  the  identity  is  true  for  all  values  of  x,  put  x  equal  to  0, 
1,  and  2  in  succession.  From  these  substitutions  the  following 
results  are  at  once  obtained : 


a;  =  0. 

1  =  2^, 

A  =  h; 

x  =  l, 

2  =-5, 

B  =  -2; 

x  =  2. 

5  =  2(7, 

C=|. 

X2+1 

1 

2      .         5 

x(x-l)(x-2)      2x     x-1      2(x-2) 

X^  4-  X^  -\-  1 

Ex.  3.     Resolve  — — — —  into  partial  fractions. 

x3-3x  +  2  . 

The  process  of  decomposition  cannot  be  applied  directly  to  this 

fraction,  because  the  numerator  is  not  of  lower  degree  than  the 

denominator.    But,  by  division,  any  such  fraction  can  be  replaced 


474  INDETERMINATE   COEFFICIENTS. 

by  an  integral  function  plus  another  fraction  whose  numerator  is 
of  lower  degree  than  the  denominator.     In  the  present  case  we  find 

a;3  +  x^  +  1  _  J      x'^  +  3  a;  -  1 
x3_3x  +  2  a;3-3x  +  2' 

and  we  apply  the  process  of  decomposition  to 

(x^  +  3x  -  1)  /(x3  -  3x  +  2). 
Since  x^-Sx  +  2=(x-  iy(x  +  2), 

this  latter  fraction,  for  auo;ht  we  as  yet  know,  might  be  the  result 
of  adding  together  three  fractions  of  the  respective  forms 

A/Cx-iy,   B/(x-l),   and    C/(x  +  2), 
for  (x—  l)2(x  +  2)    is  the   least   common   multiple   of    (x  —  1)^, 
(x  —  1),  and  (x  +  2).     Hence  we  assume 

ic2  +  3x-l_       A  B  C 


x3-3x  +  2      (x-l)2     x-1      x  +  2 
and  determine  J,  J9,  and  Cby  the  condition 

x2.4  3  X  -  1  =  ^(x  +  2)  +  ^(x  -  1)  (X  +  2)  4-  C{x  -  1)2. 
Then,  putting  x  equal  to  1,   —  2,  and  0  in  succession,  we  obtain 
the  following  sets  of  values  : 

x=l,  3  =  3^,  ^  =  1; 

a;  =  -2,    -3  =  90,  C  =  -\\ 

x  =  0,       -1  =  2^-2J5+C,    S  =  f. 

.     x3  +  x2  +  1  _  -^   ^         1         ^         4  1 


x3_3x  +  2  (x-l)2      3(x-l)      3(x  +  2) 

Should  a  factor  of  the  form  (x  + A;)^  occur  in  the  denominator 
of  the  origijial  fraction,  three  partial  fractions  ^/(x+yfc)^, 
B/{x  +  A;) 2,    G/{x  +  k)  must  be  provided  for,  and  so  on. 

2^.2  7  /v  2 

Ex.  4.    Resolve —  into  partial  fractions. 

x^  —  8 

The  denominator  of  this  fraction  may  be  resolved  into  the  two 
factors  X  —  2,  x^  +  2  x  +  4,  or  into  the  three  factors  x  —  2, 
X  +  1  +  iy/'^t  X  +  1  —  iV^?  and  observing  always  the  rule  that  the 


INDETERMINATE   COEFFICIENTS.  475 

assumed  numerator  of  any  partial  fraction  must  be  lower  in  degree, 
by  one,  than  the  denominator,  we  may  assume  either 

a;2_7x-2_    A      J  D  J  D' 


or 


a;8_8  x-2      x+l-\-  V'^     x  +  l-  iV3' 

x2-7x-2_    A      ,      5a; +C 


a;3_8  x-2x2  +  2x  +  4 

Choosing  the  latter  identity,  we  have 

x2  -  7 X  -  2  =  A(x^  -{-  2x  +  4)-\-  Bx(x  -  2)+  C(x-2), 
and  the  following  particular  equations  for  the  determination  of 
A,  B,  and  C : 

x  =  2,    -12  =  12^,  ^=-1; 

x  =  0,    -    2  =  4^-2  0,  0  =  -l; 

a;=l,     -    S  =  7A-B-C,    B  =  2. 
a;2-7x-2  1^      2x-l 


a^_8  a:-2     x^  -\-2x  +  4: 

The  calculations  for  2)  and  D'  in  the  first  identity  will  show 
that 

The  student  can  easily  verify  these  results. 

EXAMPLES   LXXXVI. 
Resolve  into  partial  fractions : 

1  5x  +  i  e      g^-f-l 

*  x2  -  1 '  '  x(x-  1)2 

2  x2  +  3 X  +  3 ,  Y       2x  +  1 

'  x2  + 3x4-2*  ■   x(x;2-f  1)* 

3  X  +  7  g       qx  -f  ft 

*  3(a^  +  3)(xH-5)*  '   x(x2  +  6)* 

^       x2^15x  +  8  g        x2  -  4 X  +  5 

(X  +  1)-^(X  -  5)*  '      (X  -  1)2(X2  +  1)' 

6  -4x  +  "B  10      7x2-llx-f  7 

(X  -  1)  (X  -  2)  (X  -  3)*  '    (x  -  l)3(x  +  2)* 


476  INDETERMINATE   COEFFICIENTS. 


11. 

5x^1 
x^-1 

16. 

^3-1 

X^+1 

12. 

4 
x*-l 

17. 

x^  +  ax+b 
x%x'^  +  b) 

13. 

2 

X(X2  +  X 

+  2) 

18. 

a;2  +  l 
x^  +  1 

14. 

X^-X 

(aj2  +  c)  (0x2  +  1) 

19. 

a:2-l 

X*  +  X2  +  1 

15. 

X2+1 

_. 

20. 

X2+1 

_. 

x3(a^  +1)  («  -  1)K«*  +  1) 


APPLICATION  TO   EXPANSION  OP  FUNCTIONS. 

333.  In  the  expansion  of  functions  into  infinite  series 
in  ascending  powers  of  the  variable,  only  such  values  of 
the  variable  are  permissible  as  will  make  the  series  con- 
vergent. For  any  other  values,  the  limit  of  the  series 
cannot  be  placed  equal  to  the  given  function,  and  if  there 
be  no  values  of  the  variable  for  which  the  series  is  con- 
vergent, the  expansion  is  impossible. 

For  example,  the  process  of  division  applied  to 
1  -7-  (1  —  £c)  produces  the  series 

l+cc-foj^-f-ic^-j ad  inf., 

but  the  limit  of  this  series  is  not  equal  to  1/(1  —  x) 
unless  abv  x<l. 

The  method  of  indeterminate  coefficients,  when  applied 
to  the  expansion  of  functions,  does  not  prevent  the 
appearance  of  divergent  series,  and  hence  the  results  of 
this  method  are  conditioned  upon  the  convergency  of  the 
series  it  produces. 


INDETERMINATE   COEFFICIENTS.  477 

Ex.  1.  Expand  (1  +  x)/{\  4-  x"^)  into  an  infinite  series. 

Assume  J-+A  -  a  +  hx  -\-  cx^  ■\-  dy?  -\- '-. 
1  +  x^ 

Then,  multiplying  by  1  +  aj^,  we  have 

1  +  x  =  a  +  &aJ  +  cx2  +  (Zx3  +  ex*  +  ••• 

+  ax2+ &x3  +  cx*  + — 

=  a  +  6x  +  (a  +  c)x2  +  (6  +  d)x3  +  -. 

for  values  of  x  for  which  the  series  is  convergent.    Hence,  by  the 
principle  of  indeterminate  coefficients  [Art.  330], 

a  =  l,       .6  =  1,  a  +  c  =  0,  h-\-d  =  0, 

c  =  — 1,        d  =  — 1,        e  =  l,       /=  1,        etc. 

.-.  J-t4=H-x-x2-x8  +  x4  4-a:^-a:« . 


Ex.  2.   Expand into  an  infinite  series. 

Decomposing  this  function  into  partial  fractions,  we  find 

2  _     1     _^  1  +  x 


-x8  +  a2-x+l      1-x     l+x2, 
But,  for  values  of  x  for  which  the  series  are  convergent, 
1 


l  +  x  +  a;''  +  a^  +  x4  +  x5  +  ... 
and  }'^\  =  1  +  X  -  x2  -  x8  +  X*  +  a^ 


1  -X 
IjfX 
1+X2 


X3  +  X2  -  X  4-  1 


2(1  +  X  +  X*  +  a;6  +  x8  +  x9  +  — )• 


Ex.  3.   Expand "^ into  an  infinite  series. 

(x  -  2)2(2  x-1) 

This  fraction  is  equal  to  the  difference 

2 1__ 

(x-2)a      l-2x* 


478  INDETERMINATE   COEFFICIENTS. 

2 

Assume =  ao  +  uix  +  a2X^  +  •-. 

(x  —  2)2 

Then  multiplying  by  4  —  4  a;  +  a^'^,  we  have 

2  =  4  ao  +  4  aix  +  4  a2X^  +  4  aga^s  +  ... 

—  4  aox  —  4  aix2  —  4  ^2x3  —  ••• 

+    aox^  -\-    aix^  H 

for  values  of  x  that  make  the  series  convergent.    Hence  ^ 

4  Gfo  =  2,         4  «i  —  4  ao  =  0,         4  02  —  4  ai  +  ao  =  0, 

and  in  general  4  a„  —  4  an-i  +  an-2  =  0. 

From  these  equations  the  values  of  ao,  ai,  «2»  •••  «n  can  he  cal- 
culated in  succession.     Thus 

1  2  3  4 

and  in  general,  an  =  — J^- 

2  1,    2^^3  4^3  +  4^4  +  ...; 


(x-2)2      2      22         23  2*  25 

and  then,  since 

— ^  =  1  +  2  X  +  22^2  +  23a;3  +  2^x'^  +  — . 
1  —  2x 

(x  -  2)2(2  x-1)      \2        I      \2^         )        \2^  I 

the  general  term  of  the  series  being  ( —  —  2"-^  J  xP--'^. 

Observe  that  for  positive  values  of  x  every  term  of  this  series  is 
negative,  but  that  the  fraction  is  not  negative  unless  x<.\. 

334.    Keversion  of  Series.     Given 

y  =  aQ-^a^x  +  a^'^-\-a^o?-\ ,  (i.) 

we  may  propose  to  obtain  a  value  of  x  in  terms  of  y. 
For  this  purpose  we  assume 

a;  =  60  +  &i2/  +  M'  +  %'  +  -,  (ii.) 


INDETERMINATE  COEFFICIENTS.  479 

and  determine  the  coefficients  b^,  bi,  bz,-"  by  the  condi- 
tion that  the  given  value  of  y  in  (i.),  when  substituted 
in  (ii.),  produces  an  identity  in  x. 

The  process  is  likely  to  be  laborious,  but  its  applica- 
tion is  sometimes  successful  and  convenient,  as,  in  Art. 
359.     The  process  is  called  the  reversion  of  series. 

Ex.  Find  a  value  of  x  that  will  satisfy  the  equation  2  x  —  ic*  =  y 
for  a  given  value  of  y. 

Even  powers  of  y  will  not  appear  in  the  result,  and  we  may 
assume 

x  =  h{y  +  b0^  +  b^y^  -\ 

=  &i(2a;-ic3)4-&3(2ic-x3)8+  .... 
Expanding  (2x  —  x^y,  (2x  — x^)^  etc.,  and  comparing  coeffi- 
cients, we  obtain 

^      4       32        64       2048 

This  solution  is  effective  for  any  value  of  y  for  which  the  series 
is  convergent. 

EXAMPLES   LXXXVII. 

Expand  the  following  functions  into  infinite  series  in  ascending 
powers  of  x  and,  when  possible,  obtain  a  formula  for  the  general 
term: 

,     5x+  1  g      l  +  a;g 

•  -•  •    (l-x)»* 


5x+l 

1-X2 

2 

1-4x4-3x2 

4-3x2- 

-2x3 

1-x- 

2x3 

1 

(1  +  xy 

1 

2.    = 7. 

1  -  4  X  4-  3  x2 

3    4-3x2-2x«  g 


1  4-a; 


x2  +  15x  +  8 


(X  +  l)2(x  -  6) 


6. — -•  10. 


1 


11.  Prove  that,  if  n,  r  =  positive  integers,  the  coefficient  of 
xn+r-i  in  the  expansion  of  (1  +  x")  /(I  -  x)2  is  (n  +  2  r). 


480  INDETERMINATE   COEFFICIENTS. 

APPLICATION  TO   SUMMATION  OF   SERIES. 

33^.  If  the  general  term  of  a  series  be  given  as  a 
rational  integral  function  of  the  number  of  terms,  the 
sum  of  n  terms  may  be  found  by  the  method  of  indeter- 
minate coefficients. 

Denoting  the  nth  term  of  the  series  by  u^,  let 

Sn  =  %  +  «^2  +  W3  H h  W„, 

Sn-\  =  Wi  +  ^2  +  ^3  H f-  ^n-lj 

then,  for  all  values  of  n, 

If,  now,  u^  be  given  in  the  form 

w„  =  tto  +  otin  4-  a2n^  -\ , 

and  we  assume  that 

^S'n  =  &o  +  bin  +  62^^  +  M^  H 

for  all  values  of  n,  where  bo,  &i,  62?  •••  ^-re  as  yet  undeter- 
mined coefficients,  then 

Sn-j  =  &0  +  bi{n  -  1)  +  b,{n  -  1)^  +  -, 
and 

S^-S^.i  =  h{n-(n-l)l  +  b,\n'-(n-iyi  +  ... 

^bi-h  b2(2n  -  1) 4-  bs(3n^ -  Sn  + 1) 

4-  64(47^3  -  67i2  +  4n  -  1)  -V  ..., 

and  we  have  the  identity 

tto  +  ttin  +  «2^^  4-  ••• 

\^  ^bi-h-^-bs 4-(2&2-3&3+-)^ 

+  (363-664+ •••)^'+-. 


INDETERMINATE  COEFFICIENTS.  481 

from  which  we  obtain  the  values  of  bi,  62?  ©tc,  by  equat- 
ing the  coefficients  of  like  powers  of  n. 

Suppose,  for  example,  that  w„  is  of  a  degree  in  n  not 
higher  than  2.     Then 

&4=&5  =  &6=  •••  =  0, 

and        &i  —  &2  +  ^3  =  <^o >  2 62  —  3  63  =  dj ,  3 63  =  a2, 
and  from  these  equations  are  derived  in  succession 

Sn  =  h  +  («o  +  !«!  +  ia2)n  +  ^(tti  4-  a2)w^  +  Jagn^ 

The  value  of  bo  may  then  be  found  by  putting  w  =  1  in 
this  formula. 

Observe  that  if  n^  be  the  highest  power  of  n  contained 
in  S„,  the  difference  S^  —  *S'„_i  will  contain  71^"^  but  not 
n^.  Hence,  the  degree  {in  n)  of  /S^  is  higher  by  1  than  the 
degree  ofu^. 

Ex.  1.  Find  the  sum  of  the  first  n  terms  of  the  series  whose  n^ 
term  is  1  —  w  +  ri^. 

Here  ao  =  1,  ai  =  —  1,  02  =  1 ; 

=  &o  +  fn  +  |n8; 

and  since  ^1  =  1  =  6©  +  1,  that  is,  60  =  0, 

...  >S„  =  }n(n2  +  2). 

Ex.  2.  Find  the  sum  of  the  first  n  terms  of  the  series  3  4-  4  + 
6  +  9  +  13  +  18+  .... 

The  differences  of  successive  pairs  of  terms  of  this  series  form 
the  arithmetical  progression  1,  2,  3,  4, ...  ;  that  is, 

4-3=1,  6-4=2,  9-6  =  3,  etc. 
2h 


482  INDETERMINATE   COEFFICIENTS. 

Hence,  denoting  the  successive  terms  of  the  given  series  by  wi, 
M2»  Us,  W4,  •••  Un,  we  have 

Ml  =  3,    W2  =  Ml  +  1, 

us  =  ui  +  l  +  2, 
M4  =  wi  +  1  +  2  +  3, 


«n  =  Wi  +  l  +  2  +  ...4-  (n-l) 

=  Wi  +  Kw-l)(2  +  w-l-l)  [Art.  259] 

=  Mi+  ^W  (W  —  1). 

This,  the  n^^  term  of  the  given  series,  is  therefore  3  —  ^w+^w^. 
Hence,  in  the  formula  for  Sn,  we  have 

ao  =  3,  ai  =  —  I,  a2  =  ^, 
5n  =  60  +  (3  -  i  +  tV)  »*  +  K-  i  +  D  w2  +  iw« 
=  60  +  ¥  »i  +  i  w^  ; 
and  since  /S'l  =  3  =  60  +  3,  bo  =  0. 

Let  the  student  verify  the  results  of  these  two  examples  by  use- 
ing  the  method  here  explained,  but  without  substituting  in  the 
formula. 

336.  The  foregoing  method  may  be  generalized  and 
made  applicable  to  power  series  in  which  the  coefficient 
of  a;**  is  a  rational  integral  function  of  n. 

For  the  summation  of  such  a  series  it  will  be  found 
sufficient,  when  the  (n  +  1)*^  term  is 

to  assume 

S,^,  =  (60  +  &in  +  •••  +  hn')x-+'  +  k 
and  determine  60?  ^\y"'^h  ^^^  ^  from  the  identity 

An  example  will  best  show  how  the  method  is  applied. 


INDETERMIKATE  COEFFICIENTS.  483 

Ex.  1.  Find  the  sum  of  the  first  w  +  1  terms  of  the  series 
1  + 2a; +  3x2 +  4x3  +  .... 

The  (w  +  l)th  terra  of  the  series  is  (n  +  1)  x",  and  if  aS^„+i  denote 
the  sum  of  the  first  w  +  1  terms,  then 

(n  +  l)x^  =  S„+i  -  Sn. 

Assume  Sn+i  =  (a  +  6w)x"+i  +  k, 

where  k  is  independent  of  n,  and  a,  b  are  as  yet  undetermined 

coefficients.    Then 

S„  =  {a-\-  bin  -  l)}x«  +  k, 

and  Sn+i-  Sn  =  {a(x-l)-\-b  +  6(x  -  l)n}x», 

whence  w  +  1  =  a(x  —  1)+  6  +  &(x  —  l)n. 

Equating  coefficients  in  this  equation,  we  now  have 

a(x-  1)+  6  =  1,  6(x-l)=l, 

K         l^l-6x-2 
or  0  = ,  a  = 


x-1  x-1      (x-l)2 

Hence  5„,,  =  {  ^  +  -^  }..+•  + *, 

for  all  values  of  n.     We  now  determine  k  by  putting  n  =  0  in  this 
expression  for  Sn+i,  and  by  this  substitution  we  obtain 

that  is.  k  = 


(X  -  1)2 
Hence  finally,     Sn+i  =  {  -^^  +  -^  }a;«+i  +  — ^, 

I  (X  —  1)-^        X—  IJ  (X  —  1)2 

which  is  the  required  result. 

EXAMPLES  LXXXVIII. 

Each  of  the  following  expressions  is  the  nth  term  of  a  series 
whose  first  term  is  got  by  putting  n  =  1.  Find  the  sum  of  the 
first  n  terms  of  each  of  the  series  thus  indicated. 

1.    2n2-l.  3.    w(n  + l)(7i  +  2).  6.  p  +  g(n  -  1). 

3.   n(n+l).  4.   n^-n  +  1.  6.    — 

4n2  —  1 


484  INDETERMINATE   COEFFICIENTS. 

Each  of  the  following  expressions  is  the  (n  +  1)''^  term  of  a 
series  whose  first  term  is  got  by  putting  n  =  0.  Find  the  sum  of 
the  first  n  +  I  terms  of  each  of  the  series  thus  indicated. 

7.  (p  +  qn^x"^.     ,  10.    (p  +  qn  +  rn^)x'^. 

8.  (w2  -  w  +  l)x^.  11.    (2'»  -  2  w  +  l)ic«. 

9.  {2{n  +  ly  -  l}x^.  J2    _wMi2jL+2_2n 

'    (n  +  l)(w  +  2) 

Find  the  sum  of  the  first  n  terms  of  each  of  the  following  series : 

13.  1 +2x4-5x2+ 10x3  + 17x4  + 26a:5+.... 

14.  1 .  2  .  X  +  2  .  3 .  x^  +  3 . 4 .  x3  +  4 .  6 .  X*  +  ••.. 

15.  1 +  2x  + 4x2  + 7x3 +  llx*  + 16x5  + .... 

16.  1  +  2x  +  6x2  +  13x3 +  23x*  + 36x5  +  .... 


PERMUTATIONS   AND   COMBINATIONS.  485 


CHAPTER  XXXIII. 

Permutations  and  Combinalions. 

337.  Definition.  The  different  ways  in  which  r  things 
can  be  taken  from  n  things,  regard  b^ing  had  to  the  order 
of  selection  or  arrangement,  are  called  the  permutations  of 
the  n  things  r  at  a  time. 

Thus  two  permutations  will  be  different  unless  they 
contain  the  same  objects  arranged  in  the  same  order. 

For  example,  suppose  there  are  four  objects,  represented  by  the 
four  letters  a,  6,  c,  d. 

If  we  take  the  four  objects  one  at  a  time,  we  have  the  four  per- 
mutations a,  &,  c,  d. 

If  we  take  the  objects  two  at  a  time,  we  have  twelve  permu- 
tations 

a?),  ac,  ad,  ba,  be,  bd,  ca,  cb,  cd,  da,  db,  dc. 

Now  to  find  the  whole  number  of  the  permutations  three  at  a 
time  we  can  proceed  thus.  Take  any  one  of  the  permutations 
two  at  a  time  and  place  after  it  either  of  the  two  letters  which  it 
does  not  contain ;  then  all  the  permutations  so  obtained  will  be 
different,  for  either  we  shall  have  used  a  different  permutation  of 
two  letters  or  else  the  final  letters  will  be  different ;  moreover, 
since  all  the  possible  permutations  two  together  have  been  used, 
no  permutation  of  three  letters  can  have  been  omitted.  Hence 
the  number  of  permutations  of  4  things  3  together  =  number  of 
permutations  of  4  things  2  together  x  2  =  12  x  2  =  24. 

We  can  by  similar  reasoning  find  the  permutations  of  n  things 
r  together. 


486  PERMUTATIONS    AND   COMBINATIONS. 

The  number  of  permutations  of  n  different  things 
taken  r  at  a  time  is  denoted  by  the  symbol  ^P^. 

338.  To  find  the  number  of  permutations  of  n  different 
things  taken  r  at  a  time,  where  r  is.  any  integer  not  greater 
than  n. 

Let  the  different  things  be  represented  by  the  letters 
a,  b,  c, '-'. 

It  is  obvious  that  there  are  n  permutations  of  the  n 
things  when  taken  one  at  a  time,  so  that  „Pi  =  n. 

Now,  if  beside  any  one  of  the  different  permutations 
(r  —  1)  at  a  time,  we  place  any  one  of  the  n  —  (r  —  1) 
letters  which  it  does  not  contain,  we  shall  obtain  a  per- 
mutation of  the  n  things  r  at  a  time.     We  thus  obtain 

n  —  {r  —  l)  =  n  —  r-\-  1 
different  permutations  r  at  a  time  from  every  one  of  the 
different  permutations  (r  —  1)  at  a  time. 

Hence  „P,  =  „P,li  x  (n  -  r  +  1). 

Since  the  above  relation  is  true  for  all  values  of  r,  we 
have  in  succession 

nPr-l=nPr-2Xin-r  +  2), 
nPr-2=nPr-sX{n-r-h3), 

,P3  =  nAx(il-2), 

Also  „Pi  =  n. 

Multiply  all  the  corresponding  members  of  the  above 
equalities,  and  cancel  all  the  common  factors ;  then 

„P,  =  7i(n-l)(n-2)...(7i-r  +  l), 
r  being  the  number  of  factors  on  the  right. 


PERMUTATIONS   AND   COMBINATIONS.  487 

If  all  the  n  things  are  to  be  taken,  r  is  equal  to  n,  and 
we  have 

„P„  =  n(7i-l)(7i-2)...3.2.1. 

Def.  The  continued  product  71(71  —1 )  (n  —  2)  ...  3  •  2  •  1 
is  denoted  by  the  symbol  \n,  or  by  n  !  The  symbols  \n 
and  n !  are  read  '  factorial  n.'     Thus  [4  =  4 !  =4  .  3  .  2  •  1. 

Ex.  1.  In  how  many  different  ways  can  6  boys  stand  in  a  row  ? 
The  number  required  =  ePe  =  6  •  5  •  4  •  3  •  2  •  1  =  720. 

Ex.  2.  How  many  different  numbers  can  be  formed  by  using 
three  of  the  figures  1,  2,  3,  4,  5  ? 

The  number  required  =  6P3  =  5  •  4  •  3  =  60. 

Ex.  3.  Show  that  10P4  =  7P7, 

10P4  =  10  -9 .  8  .  7  and  7P7=  7  ..6 . 5 •  4 . 3  .2  . 1. 

339.  To  find  the  number  of  permutations  of  n  things 
taken  all  together,  when  the  things  are  not  all  different. 

Let  there  be  n  letters ;  and  suppose  p  of  them  to  be 
a's,  q  of  them  to  be  5's,  r  of  them  to  be  c's,  and  so  on. 

Let  P  be  the  required  number  of  permutations. 

If  in  any  one  of  the  actual  permutations,  we  suppose 
that  the  a's  are  all  changed  into  p  letters  different  from 
each  other  and  from  all  the  rest;  then,  by  changing  only 
the  arrangement  of  these  jp  new  letters,  we  should,  instead 
of  a  single  permutation,  have  p !  different  permutations. 

Hence,  if  the  a's  were  all  changed  into  p^  letters  differ- 
ent from  each  other  and  from  all  the  rest,  the  6's,  c's,  etc., 
being  unaltered,  there  would  be  P  xpl  permutations. 

Similarly,  if  in  any  one  of  these  new  permutations  we 
suppose  the  6's  are  all  changed  into  q  letters  different  from 
each  other  and  from  all  the  rest,  we  should  obtain  q !  per- 
mutations by  changing  the  arrangement  of  these  q  new 


488  PERMUTATIONS   AND   COMBINATIONS. 

letters.    Hence  the  whole  number  of  permutations  would 
now  be 

Pxpl  xql 

By  proceeding  in  this  way  we  see  that  if  all  the  letters 
were  changed  so  that  no  two  were  alike,  the  total  number 
of  permutations  would  be 

Pxpl  xql  xr\  .... 
But  the  number  of  permutations  of  n  different  things  is  n !. 
Hence  P xpl  X  ql  X  rl  x  -•-  =  nl; 

nl 


.-.    P: 


plqlrl 


Ex.  1.  Find  the  number  of  permutations  of  the  six  letters 
aaabbc  taken  all  together. 

The  number  =  — — —  =  60. 
3!2! 1! 

Ex.  2.  How  many  different  numbers  can  be  formed  by  the 
figures  2,  3,  3,  4,  4,  4,  4  ? 

7  ' 

The  number  required  = '—  =  106. 

^  4!2! 

\ 

340.  Def.  The  different  ways  in  which  a  selection 
of  r  things  can  be  made  from  n  things,  without  regard 
to  the  order  of  selection  or  arrangement,  are  called  the 
Oombinations  of  the  n  things  r  at  a  time. 

Thus  two  combinations  will  be  different  unless  they 
both  contain  precisely  the  same  objects. 

For  example,  suppose  there  are  four  objects,  represented  by  the 
four  letters  a,  6,  c,  d.  The  different  combinations  one  at  a  time 
are  a,  b,  c,  d  ;  the  combinations  two  at  a  time  are  ab,  ac,  aj?,  6c, 
bdy  cd  ;  the  combinations  three  at  a  time  are  abc,  ubd,  acd,  bed ; 


PERMUTATIONS   AND   COMBINATIONS.  489 

and  there  is  only  one  combination,  namely  ahcd^  when  aU  the 
letters  are  taken. 

The  number  of  combinations  of  n  things  r  together  is 
denoted  by  the  symbol  „0^. 

341.  To  find  the  number  of  combinations  of  n  different 
things  taken  r  at  a  time. 

Since  every  combination  of  r  different  things  would 
give  rise  to  r\  different -permutations,  if  the  order  of  the 
letters  were  altered  in  every  possible  way,  it  follows  that 

„axr!  =  „P,  =  n(ii-l)(n-2)...(n-r  +  l). 

Hence    ^o^^^(^- 1)(^- 2)-(^ -^+ 1). 
1.2.3...r 

By  the  following  method  the  number  of  combinations 
of  n  different  things  r  at  a  time  can  be  found  indepen- 
dently of  the  number  of  permutations. 

Let  the  different  things  be  represented  by  the  n  letters 
a,  &,  c,  d,  .••.  4  ^v 

In  the  combinations  of  the  n  letters  r  together  the 
number  in  which  a  particular  letter  occurs  must  be  equal 
to  the  number  of  ways  in  which  r  —  1  of  the  remaining 
n  —  1  letters  may  bif5(pelected.  Hence  in  the  whole 
number  of  combinations  r  together  every  letter  occurs 
n-\Cr-i  times,  and  therefore  the  total  number  of  letters 
employed  is  n-\Gr-i  X  n;  but,  since  there  are  r  letters  in 
each  of  the  „0^  combinations,  the  total  number  of  letters 
employed  is  „C^  x  r. 

Hence  r  x  ,,(7^  =  ?i  X  „_iC^_i. 

Since  the  above  relation  is  true  for  all  values  of  n  and 
r,  we  have  in  succession 


490  PERMUTATIONS   AND   COMBINATIONS. 

(r -  1)  X  ,  ia_i  =  {n  -  1)  X  ._2a-2 

(r  -  2)  X  n-20r-2  =  {u  -  2)  X  n-A-^ 


2  X  „_,+2C2  =  (n  -  r  +  2)  X  n-.+iCi. 
Also  n-r+iCi  =  n  —  r  4- 1. 

Multiplying    corresponding    members    of    the    above 
equalities  and  cancelling  the  common  factors,  we  have 

r !  X  „a  =  71  (71  -  1)  (n  -  2) . ..  (n  -  r  +  1), 

that  is  .0.  =  ri(n-l)(n-2)...(n-r4-l), 

r ! 

By  multiplying  the  numerator  and  denominator  of  the 
fraction  on  the  right  by  {n  —  r) ! ,  we  have 

r  _^(y^-l)(^-2)-»-(n-r  +  l)(n  — r)!_  *    n 


r\{n  —  r)\  r\{n—r)\ 

Ex.  1.   How  many  groups  of  3  boys  are  there  in  a  class  of  15  ? 

The  number  =  15C3  =  ^^'^^'^^  =  455. 
1  •  2 • 3  ^^ 

Ex.  2.  There  are  6  candidates  for  4  vacancies,  and  every 
elector  can  vote  for  any  number  of  candidates  not  greater  than 
the  number  of  vacancies.  In  how  many  ways  is  it  possible 
to  vote  ? 

An  elector  can  vote  for  4  candidates  in  ^d  ways,  for  3  candi- 
dates in  eCs  ways,  for  2  candidates  in  6  02  ways,  and  for  1  candidate 
in  eCi  ways.  Thus  the  whole  number  of  ways  in  which  an  elector 
can  vote  is 

6C4  +  6C3  +  6^2  +  eCi  =  16  +  20  +  15  +  6  =  66. 


( ' 


PERMUTATIONS   AND  CpifBiNATIONS.  491 

342.  From  the  formula  obtained  in  the  last  article^ 
we  have 

r     -       ^- 

{n  —  r)\r\ 

I  SO  that  the  number  of  combinations  of  n  things  r  together 
is  equal  to  the  number  of  combinations  of  n  things  n  —r 
together. 

The  above  proposition  follows,  however,  at  once  from 
the  fact  that  whenever  we  take  r  out  of  n  things  we 
must  leave  n  —  r,  and  if  every  set  of  r  things  differs  in 
some  particular  from  every  other,  the  corresponding  set 
of  n  —  r  things  will  also  differ  in  some  particular  from 
every  other.  Hence  the  number  of  different  sets  of  r 
things  must  be  equal  to  the  number  of  different  sets  of 
n  —  r  things. 

343.  To  prove  that  JO,  +  „C,_i  =  nH-i<^r- 
The  total  number  of  combinations  of  (71  + 1)  things  r 

together  can  be  divided  into  two  groups  according  as  they 
do  or  do  not  contain  a  certain  particular  thing.  The 
number  which  do  not  contain  that  particular  thing  is 
the  number  of  ways  in  which  r  of  the  remaining  n  things 
can  be  taken,  which  is  „(7^;  and  the  number  which  do 
contain  the  particular  thing  is  the  number  of  ways  in 
which  (r  —  1)  of  the  remaining  n  things  can  be  taken, 
which  is  J3r-\'     Hence  ,,+iC^  =  J^,  +  rfir-v 

Thus,  if  n  =  7  and  r  =  4,  we  have 

7-6.5-4-3-2.1^6-5.4-3.2>l      6..5.4.3-2-1 
4. 3. 2. 1-3. 2-1.     4-3.2. 1-2.1      3.2.1.3.2-^' 


492  PERMUTATIONS   AND   COMBINATIONS. 

Note.  —  The  above  important  proposition  may  also  be  proved 
from  the  general  formula.     Thus 

vf^r  +  n^r— 1 

_  n{n  -V){n-  2)...(n  -  r  +  1>     n{n  -  l)(n  -  2)  — (w  -  r  +  2) 
1.2.3...r  1.2.3...(r-l) 

^  n{n  -  l)(n  -  2). -(71  ->  +  2)  ..  . 

1.2.3...r  ^^  -r    y-r  5 

_  (w  +  l)w(n  -  l)(n  -  2)---(n  -7-4-2) 
1.2.3...»- 

344.  Greatest  Value  of  „C7^ 
T'rom  Art.  341  we  have 

y^n  —  In  —  2  ^yi-r  +  1 

.     n  -   n      y n-r  +  1 
r 

Hence  rflr^nCr-i  according  as '^^—  r.  1 ;  that  is, 

according  as  r  ^  i  (n  +  1) . 

Thus  the  number  of  combinations  of  n  things  increases 
as  r  increases,  st)  long  as  r  is  less  than  -^-(w  +  l).  If 
r  =  ^{n  4- 1),  then  ^Cr  =  nC^-u  but  r  cannot  be  equal  to 
1(91  +  1)  unless  n  is  odd.  If  r  >  i(w  +  1),  the  number 
of  combinations  diminishes  as  r  increases. 

Thus,  if  n  is  even,  „C^  is  greatest  when  r  —  ^w,  and  if 
n  is  odd,  „(7^  =nC'r-i  when  r  =  |^(n  +  l),^and  these  values 
are  the  greatest. 

Ex.  1.   Find  the  greatest  value  of  gCV- 
^      The  greatest  value  is  when  r  =  4,  and  the  value  is 
8  .  7  »  6  •  5  ^  ^Q 
^  1.2.3.4 


PERMUTATIONS   AND  COMBINATIONS.  493 

Ex.  2.   Find  the  greatest  value  of  ^iCr. 

The  greatest  values  are  when  r  =  6  and  r  =  5,  and  the  values  are 

ii.io.9.8.7^^ea 

1.2.3.4.5 


EXAMPLES  LXXXIX. 

1.   Find^sPg.  2.   FindioP^.  3.   Find  gP^. 

4.    How  many  different  numbers  can  be  formed  by  using  the 
five  figures  1,  2,  3,  4,  5? 

6.   How  many  different  numbers  can  be  formed  by  using  one  or 
more  of  the  figures  1,  2,  3,  4,  5,  6  ? 

6.    In  how  many  ways  can  10  boys  be  put  in  a  row  so  that  two 
particular  boys  should  not  be  together  ? 

1/       7.   Of  how  many  different  things  are  there  720  permutations 
when  taken  all  together  ? 

8.   Find  the  number  of  permutations  of  all  the  letters  of  each 
of  the  words  success,  Mississippi,  and  algebraic. 

1  9.  In  how  many  ways  can  the  nine  letters  aaaabbbcc  be  ar- 
ranged in  a  row  ? 

\  10.  How  many  different  numbers  can  be  formed  by  using  the 
seven  figures  1,  1,  1,  2,  ^,  3,  3  ? 

^  11.  How  many  permutations  can  be  made  with  the  letters  of  the 
word  essences,  and  how  many  begin  with  e  and  end  with  s  ? 

I     12.   Find  ^Pn,  having  given  that  nPg  =  nP^  x  2. 

13.  Find  n,  having  given  that  „Pg  =  nPj  x  12. 

14.  Find  ^^C^,  15O12,  and  ^^0^^. 

15.  JKnCs  3 

16.  If  ^,o_ 

17.  How  many  different  numbers  of  three  figures  could  be  made 
liy  taking  three  of  the  digits  0,  1,  2,  3,  4  ? 


494  PERMUTATIONS   AND   COMBINATIONS. 

18.  If  2nC'3  =  nCa  X  12,  find  n. 

19.  If  2nC,  =  nCi  X  24,  find  n. 

j  20.  Numbers  are  formed  by  writing  the  five  figures  1,  2,  3,  4,  5 
in  every  possible  order.  How  many  of  these  numbers  are  greater 
than  23,000  ? 


J 


21.    In  a  certain  town  there  are  five  letter  boxes.     In  how  many 
ays  can  a  person  post  three  letters  ? 

(  22.    How  many  different  sums  could  be  made  with  2  pennies, 
3  shillings,  and  4  sovereigns  ? 

I     23.    How  many  different  sums  could  be  made  with  4  pennies, 
6  shillings,  and  5  sovereigns  ? 

24.  Show  that  n+iCr+l  =  n^+l  +  2  nCr  +  nCr-l- 

25.  Show  that  in  gC^  the  number  of  combinations  in  which  a 
particular  thing  occurs  is  equal  to  the  number  in  which  it  does  not 
occur. 

26.  Show  that  in  ^^C^  the  number  of  combinations  in  which  a 
particular  thing  occurs  is  one-third  of  the  whole  number  of  the 
combinations. 

27.  Show  that  in  3„Cn  the  number  of  the  combinations  in  which 
a  particular  thing  occurs  is  one-third  of  the  whole  number  of  the 
combinations. 

28.  There  are  10  candidates  for  6  vacancies  in  a  committee  ; 
in  how  many  ways  can  a  person  vote  for  6  of  the  candidates  ? 

29.  At  an  election  there  are  5  candidates  and  3  members  to 
be  elected,  and  an  elector  may  give  1  vote  to  each  of  not  more 
than  3  candidates  ;  in  how  many  ways  can  an  elector  vote  ? 

30.  In  how  many  ways  can  a  picket  of  3  men  and  an  officer  be 
chosen  out  of  a  company  of  80  men  and  3  officers  ? 

31.  In  how  many  w-ays  can  a  cricket  eleven  be  chosen  out  of 
30  players  ;  and  in  how  many  different  ways  could  two  elevens  be^ 
chosen  to  play  a  match  with  one  another  ? 


PERMUTATIONS    AND   COMBINATIONS.  495 

.  -;  32.   Out  of  9  children,  of  whom  5  are  boys  and  4  girls,  in  how 
X  many  ways  can  4  be  chosen,  2  being  boys  and  2  girls  ? 

33.  In  how  many  ways  can  2  ladies  and  2  gentlemen  be  chosen 
to  make  a  set  at  lawn  tennis  from  a  party  of  4  ladies  and  6  gentle- 
men? 

34.  In  how  many  ways  can  8  children  form  a  ring  ? 

y       35.   There  are  n  points  in  a  plane,  no  three  of  which  are  in  the 
'"same  straight  line  ;  and  the  points  are  joined  in  pairs  by  straight 
lines  which  are  produced  indefinitely.     How  many  straight  lines 
are  there,  and  how  many  triangles  are  formed  by  them  ? 


/ 


496  THE   BINOMIAL   THEOREM. 


CHAPTER  XXXIV. 

The  Binomial  Theorem. 

345.  The  binomial  theorem  for  a  positive  integral 
index  was  proved,  by  induction  in  Art.  205.  The  follow- 
ing proof  makes  use  of  the  principles  of  the  combinatorial 
analysis,  explained  in  Chapter  XXXIII. 

Suppose  we  have  n  factors,  each  of  which  is  a  +  6.  If 
we  take  a  letter  from  each  of  the  factors  of 

(a  +  6)(a  +  6)(a  +  6)-" 

and  multiply  them  all  together,  we  shall  obtain  a  term  of 
the  continued  product ;  and  if  we  do  this  in  every  possible 
way  we  shall  obtain  all  the  terms  of  the  continued  product. 

Now  we  can  take  the  letter  a  every  time,  and  this  can 
be  done  in  only  one  way,  hence  a**  is  a  term  of  the 
product 

The  letter  b  can  be  taken  once,  and  a  the  remaining 
(n  —  1)  times,  and  the  number  of  ways  in  which  one  h 
can  be  taken  is  the  number  of  ways  of  taking  1  out  of  n 
things,  so  that  the  numbet  is  „(7i;  hence  we  have  a 
second  term 

Again,  the  letter  h  can  be  taken  twice,  and  a  the 
remaining  {n  —  2)  times,  and  the  number  of  ways  in 
which  two  6's  can  be  taken  is  the  number  of  ways  of 


THE  BINOMIAL  THEOREM.  497 

taking  2  out  of  n  things,  so  that  the  number  is  ^Cg;  hence 
we  have  a  third  term 

And,  in  general,  b  can  be  taken  r  times  (where  r  is  any 
positive  integer  not  greater  than  n)  and  a  the  remaining 
(n  —  r)  times,  and  the  number  of  ways  in  which  r  6's 
can  be  taken  is  the  number  of  ways  of  taking  r  out  of  n 
things,  so  that  the  number  is  ^C^ ;  hence  we  have,  as  the 
(r+l)*^term,  ^^^-^ 

The  letter  h  can  be  taken  every  time  in  only  one  way ; 
hence  we  have  the  term  &",  which  agrees  with  the  result 
obtained  by  putting  r  =  ?i  in  „C^  •  a"  '"6'*,  since  „(7„  =  1. 

Thus  {a-\-h){a  -\-h)  {a -\-h)  "•  to  n  factors 

=  a"  +  nCi  •  a"-^6  +  nC2a"  ^6=^  +  •••  +  nC^a^'-'-ft"  +  ...  +  6«. 

Hence,  when  n  is  any  positive  integer,  we  have 
{a+by  =  a«  +  nOi  •  a'^-'b  +  A  •  «""'^'  -}-••• 

This  proves  the  binomial  theorem  for  a  positive 
integral  index. 

The  series  on  the  right  is  called  the  expansion  of 
{a  +  by. 

If  in  this  we  Substitute  the  known,  values  [Art.  341] 
of  nGi,  „C2,  etc.,  we  obtain  the  form  in  which  the  theorem 
was  written  in  Art.  205,  namely : 

(a  +  6)"  =  a^  +  na''-'b  +  '^i'^-^)  a^-^b-  +  ... 


2i 


498  THE   BINOMIAL   THEOREM. 

346.  General  Term.  Any  term  of  the  expansion  oi 
(a  +  6)'*  will  be  found  by  giving  a  suitable  value  to  r  in 

^(^  _  1)  (n  -  2)  . ..  (n  -  r  +  1)  ^n-rj^r^ 

rl 

This  is  called  the  general  term  of  the  expansion.  It 
should  be  noticed  that  it  is  the  {r-\-iy^  from  the  beginning. 

347.  Some  applications  of  the  binomial  theorem : 
If  we  put  w  =  2  in  the  binomial  formula,  we  have 

(a  +  6)2  ^  «2  +  2  ab  +  b^. 
If  we  put  n  =  3,  we  have 

(a  +  by  =  a^  +  ^a'^b  +  — «62  +  js 

1  1   •   iS 

=  a^-{-Sa^b  +  Sab^-\-bK 
If  we  put  w  =  4,  we  have 

(a  +  by  =  a*  +  ^a^ft  +  f^a^S^  +  f44«&'  +  ^* 
1  x  •  2  1  •  2  •  o 

=  a*  +  4  a36  +  6  a^-b^  +  4  aft^  +  b\ 
If  we  put  2 X  for  a,  and  —3y  for  b,  and  «  =  5,  we  have 
(2x-Syy=(2xy  +  6(2xy(-Sy)  +  ^(2xy(i-Syy 

+f^(2a^)^(-3y)«  +  ^;^;^;^(2x)(-y)H(-3y)5 
=  32x5  -  240  a;*?/  +  720  ic^  _  io80  x'V  +  SlOiicy*  _  243^5. 

348.  In  the  expansion  of  (a +  6)"  by  the  binomial 
theorem,  the  (r  + 1)*^  term  from  the  beginning  and  the 
(r-\-iy^  term  from  the  end  are  respectively 

^CrCi'^'-'b'  and  JJ^_,a'}f-\ 

But  „C,  =  „0„_, ,  for  all  values  of  r.     [Art.  342.] 


THE   BINOMIAL  THEOREM.  499 

Hence  in  tlie  expansion  of  (a +  6)'',  the  coefficients  of 
ayiy  tivo  terms  equidista^it  from  the  beginning  and  the  end 
are  the  same,  so  that  the  coefficients  in  order  are  the  same 
when  read  backwards  as  when  read  forwards. 

349.  If  in  the  formula  of  Art.  345  we  put  a  =  1  and 
b  =  x,  we  have 

1-2  rl{n—r)\ 

This  is  the  most  convenient  form  of  the  binomial  theo- 
rem, and  the  one  which  is  generally  employed. 

It  should  be  noticed  that  the  above  form  of  the  bino- 
mial theorem  includes  all  cases;  if,  for  example,  we 
want  to  find  (a  4-  6)",  we  have 

(a  +  6)"=  {  a(l  + 1)  }"=  a"(l  +  ^)"=  a'(l  +  «^  +  etc.) 

=  a"  +  wa"-^6  -f  etc. 

350.  Greatest  Term.  In  the  expansion  of  (l  +  x)"  by 
the  binomial  theorem,  the  (r-f-1)"^  term  is  formed  from 
the  r**^  by  multiplying  hj  x(n  —  r  +  1)/^. 

Now  x(n  —  r-\-  l)/r  =  x\(n-\-  l)/r  —  Ij,  and  (n-\-l)/r 
clearly  diminishes  as  r  increases;  hence  x{n  —  r-\-l)/r 
diminishes  as  r  is  increased.  If  a;(n  — r  +  l)/r  be  less 
than  1  for  any  value  of  r,  the  (r  + 1)!**  term  will  be  less 
than  the  r^\  In  order,  therefore,  that  the  r*^  term  of 
the  expansion  may  be  the  greatest,  we  must  have 


n  —  r  +  1      ^        ,      n  —  r  —  l-\-l 

X <  1,  and  X ^ >  1. 

r  '  r  —  1 

(n-[-l)x         ,     '     (n-\-l)x     ^ 
Hence    r>-^-^-p^,  and  r<\_^(    +1. 


500  THE   BINOMIAL   THEOREM. 

If  r  =  (n  4-  l)x/{n  +  1),  then  x(n  — r -{-l)/r=  1,  and 
there  is  no  one  term  which  is  the  greatest,  but  the  r'^ 
and  (r  4- 1)*^  terms  are  equal  and  are  greater  than  any 
other  terms. 

The  absolute  values  of  the  terms,  in  the  expansion  of 
(1  +  a;)*  will  not  be  altered  by  changing  the  sign  of  x; 
and  hence  the  r*^  term  of  (1  —  x)"  will  be  greatest  when 
the  r^^  term  of  (1  -|-  a;)"  is  greatest. 

Ex.  1.  .Find  the  greatest  term  in  the  expansion  of,  (1  +  x)!'' 
when  x  =  ^. 

The  r^^  term  is  the  greatest  if  r>J^  and  r<i^^  +  l.  Hence 
the  third  term  is  the  greatest. 

Ex.  2.  Find  the  greatest  term  in  the  expansion  of  (2  +  3xy^ 
when  x  =  ^. 

(2  +  Sxy^=  {2(1  +  |a:)P=  2i2(l  +  ?^xy^. 

Hence  the  f^  term  is  the  greatest  if  r  >  -3/-  and  r  <  ^^~  +  1.  Hence 
the  sixth  term  is  the  greatest. 

The  greatest  coefficient  of  a  binomial  expansion,  i.  e.  the 
greatest  value  of  ^C^,  was  found  in  Art.  344. 

EXAMPLES  XO. 
1.    Write  out  the  expansions  of : 

M'^e+ij-    f(«oe^H-i)'-(..-i)'. 

/      2.   Find  the  11th  term  in  the  expansion  of  (4  a; ]    • 

,    /  1\20 

I    3.   Find  the  17th  term  in  the  expansion  of  /  a;2 \    . 

(/73\  2n 
CC2 \      . 


THE   BINOMIAL   THEOREM.  501 


5.    Write  down  the  twq  middle  terms  in  tlie  expansion  of 

1  >^  2n+l 


^+-, 


6.  Find  the  term  in  the  expansion  of  (a  +  xy  which  has  the 
greatest  coefficient.  _^ 

7.  Find  tlie  terms  in  the  expansion  of  ^^^aJ)^  which  have  the 
greatest  coefficients.  ^B 

8.  Find  the  greatest  term  in  the  expansion  of  (1  +  2  x)^  when 
x  =  i. 

9.  Find  the  greatest  term  in  the  expansion  of  (1  +  3x)i8  when 
x  =  l 

10.  Find  the  greatest  term  in  the  expansion  of  (3  +  4  xy^  when 

11.  There  are  two  consecutive  terms  in  the  expansion  of 
(5  X  +  7)2*  which  have  the  same  coefficient :    which  are  they  ? 

12.  Write  down  the  coefficient  of  iC"  in  the  expansion  of 
(ax  -  6)". 

13.  Sliow  that  the  coefficient  of  x**  in  the  expansion  of  (1  +  a;)^'* 
is  double  the  coefficient  of  x"  in  the  expansion  of  (1  —  x)2«-i. 

14.  Show  that  the  middle  term  of  (1  +  x)2»  is 

1.3.6...(2n-l)o.^ 
nl 
I     16.  Employ  the  binomial  theorem  to  find  99*  and  999^ 

351.  Properties  of  the  Coefficients.  We  now  proceed  to 
consider  some  properties  of-  the  coefficients  of  a  binomial 
expansion.     For  this  purpose  we  write  it  in  the  form 

(1  +  xy  =  Co  +  c^x  +  c^+  '"  +  cX  +  •••  +  c^a;**, 

where 

nl 

Co  =  c„  =  1,  Ci  =  c«_i  =  n,  and  c^  =c„_^  =  — - — - — -. 

ri{n  —  r)! 

I    Put  ic=  1  in  the  above  formula;  and  we  have 
2"  =  CoH-Ci  +  C2+  —  +c^+---c». 


602  THE  BINOMIAL  THEOREM. 

Thus  the  sum  of  the  coefficients  in  the  expansion  of  (1  +  x)" 
is  2". 

II.  Again,  put  a;  =  —  1 ;  and  we  have 

( 1  -  1 )  '^  =  Co  -  Ci  H-  C2  -  Cg  +  •  •  • ; 
.-.  0  =  (Co  +  C2  4-  C4  +  •••  )  -  (ci  +  C3  +  cs  4-  •••)• 
Thus  the  sum  of  the  coefficients  of  the  odd  terms  of  a  bino- 
mial expansion  is  equal  to  the  sum  of  the  coefficients  of  the 
even  terms. 

III.  Since  c^  =  c„_^  [Art.  342],  we  may  write  the  bino- 
mial theorem  in  either  of  the  following  ways  : 

(l  +  xy  =  Co  +  CiX-^c^-\ h  cX  H c^x% 

or  (l^-a;)"=c„+c„_la;+c„_2a;24-...c„_X^-•••^-Cla;"~^+Coa;^ 

The  coefficient  of  mf"  in  the  product  of  the  two  series 
on  the  right  is  equal  to 

Co'  +  Ci2  4-C22  +  -4-c„2. 

Hence*  Cq^  +  Ci^  H +  c^^  -] cj  is  equal  to  the  coeffi- 
cient of  x""  in  (1  +  xy  X  (1  +  xy,  that  is  in  (1  +  xy^",  and 

this  coefficient  is 


(2n)\ 


nln\ 
Hence  the  sum  of  the  squares  of  the  coefficients  in  the 

expansion  of  (1  -\-xy  is  equal  to  ^ ^—-     < 

n\n\ 

Ex.  1.    Show  that 

ci  +  2  C2  +  3  cs  +  •••  +  rcr  +  '"  +  wc„  =  n2»-\ 

*  See  Art.  328. 


THE  BINOMIAL   THEOREM.  503 

Cl  +  2  C2  +  3  C3  4-  •••  +  rcr  +  •••  +  ncn 

1.2  1.2.3  ^'^r!(w-r)! 

\    ^^        ^^  1.2  (r-l)!(n-r)!  / 

=  n(l  +  l)«-i 
=  n2«-i. 

Ex.  2.  Show  that 

Co-^Ci  +ic2  -  ...  +  (-  l)n-^!!_-^_ 


2         3  n+ln+1 

2       ^3      1.2  4  1.2.3  ^       ^  n+1 

-   ^         ^    (1-1)"+1 


n+ 1      n+l 

1 
n  +  1* 

352.    To  /lid  ^Ae  continued  product  of  n  binomial  factors 

of  the  form  x-\-  a,  x  -\-h,  x  +  c,  etc. 

It  will  be  convenient  to  use  the  following  notation : 
Si  is  written  for  a  +  6  +  c  +  .••,  the  sum  of  all  the  let- 
ters taken  one  at  a  time.  /S'gis  written  for  ab-{-ac-\-bc-\ — , 
the  sum  of  all  the  products  which  can  be  obtained  by 
taking  the  letters  two'  at  a  time.  And,  in  general,  S^  is 
written  for  the  sum  of  all  the  products  which  can  be 
obtained  by  taking  the  letters  r  at  a  time. 


504  THE   BINOMIAL   THEOHEM. 

Now  if  we  take  a  letter  from  each  of  the  binomial 
factors  of 

{x  -\-a)(x-\-  b)  (x -\- c)  (x -\- d)  "  -, 

and  multiply  them  all  together,  we  shall  obtain  a  term  of 
the  continued  product ;  and,  if  we  do  this  in  every  pos- 
sible way,  we  shall  obtain  all  the  terms  of  the  continued 
product. 

We  can  take  x  every  time,  and  this  can  be  done  in  only 
one  way ;  hence  x^  is  a  term  of  the  continued  product. 

"We  can  take  any  one  of  the  letters  a,  b,  c,  •  •  •,  and  x 
from  all  the  remaining  n  —  1  binomial  factors ;  we  thus 
have  the  terms  ax''-\  bx''~\  ca;'*"^  and  on  the  whole 
^1 .  x^-\ 

Again,  we  can  take  any  two  of  the  letters  a,  b,  c,  •••, 
and  X  from  all  the  remaining  n—2  binomial  factors ;  we 
thus  have  the  terms  abx^'"^,  acx""'^,  etc.,  and  on  the  whole 
8, .  x^-\ 

And,  in  general,  we  can  take  any  r  of  the  letters  a,  6, 
c,  •••,  and  X  from  all  the  remaining  n  —  r  binomial  fac- 
tors ;  and  we  thus  have  S^  •  a?""''. 

Hence        {x-\-a){x-[-b){x-{-c){x-\-d) "' 

=  a;"  +  ^1  •  x^-^  +  Si '  aj"-2 H \-S^' aj"-*"  +  -.., 

the  last  term  being  abed---,  the  product  of  all  the  letters 
a,  b,  c,  d,  etc. 

By  changing  the  signs  of  a,  b,  c,  etc.,  the  signs  of  S^^  S^, 
/S's,  etc.  will  be  changed,  but  the  signs  of  S2,  S^,  Sq,  etc, 
will  be  unaltered.     Hence  we  have 

{x  —  a)  {x  —  b)  {x  —  c)  (x  —  d) '" 

=  a;"  -  aSi  .  a^"-^ -f /Sa  •  aj«-2  - /Sg .  a;«-3  +  ... 
+  (-  l)*-^,  .»"-'•  H h  (-  l)"a6cd .... 


THE  BINOMIAL   THEOREM.  505 

353.   The  Multinomial  Theorem.     The  expansion  of 

can  be  found  by  the  method  of  Art.  345. 
We  know  that  the  continued  product 

(a  + 6  +  C  + •••)(«  + ^  +  c4----)(ot  +  &  +  c +•••)••• 
is  the  sum  of  all  the  partial  products  which  can  be 
obtained  by  multiplying  any  term  from  the  first  multi- 
nomial factor,  any  term  from  the  second,  any  term  from 
the  third,  etc.  If  there  are  n  of  the  multinomial  factors, 
every  term  of  the  required  expansion  must  be  of  the  n^^ 
degree,  and,  therefore,  all  the  terms  must  be  of  the  form 
a'^b'd  '",  where  each  of  r,  s,  t,  etc.  is  zero,  or  a  positive 
integer,  and  r  +  s+^-f".=n. 

Now  the  term  O'd'cf  •••  will  be  obtained  by  taking 
a  from  any  r  of  the  n  multinomial  factors,  which  can  be 
done  in  „CV  different  ways ;  then  taking  b  from  any  s  of 
the  remaining  ?i  —  r  factors,  which  can  be  done  in  „_^(7<, 
different  ways ;  then  taking  c  from  any  t  of  the  remain- 
ing n  —  r  —  s  factors,  which  can  be  done  in  „_^_gO,  differ- 
ent ways ;  and  so  on.  Hence  the  total  number  of  ways 
in  which  the  term  (X''6V  •  •  •  will  be  obtained,  that  is,  the 
coefficient  of  the  term  in  the  required  expansion,  must  be 

that  is, 

n\        ^^      (n-r)!     ^^      {n-r-s)l      ^^      ^      n\ 
rl{n—r)l     s\{n—r—s)l     tl{n—r—s—t)l  rlsltl---' 

Hence  the  general  term  in  the  expansion  of 
{a  +  6  +  c-f  '"Y  is 

— — -— — a'^b'^, 
rlsltl'" 


606  THE  BINOMIAL  THEOREM. 

where  each  of  r,  s,  t,  etc.  is  zero  or  a  positive  integer,  and 
r  +  s-\-t-\ =  n. 

Ex.  1.   Find  the  coefficient  of  abc  in  (a  +  6  +  c)^. 

Here  n  =  3,  r  =  s=^t  =  l.     Hence  the  required  coefficient  is 

Ex.  2.  Find  the  coefficients  of  a*6,  a^b^,  and  a^b'^c  in  (a+fe+c)^. 
We  have  the  terms 

^a%   -^aW,   and  -^l—a%H. 
4!1!  3!2!  2!2!1! 

Hence  the  required  coefficients  are  respectively  5,  10,  and  30. 


EXAMPLES  XCI. 

In  the  following  examples  co,  Ci,  C2,  are  the  coefficients  of  jc^, 
a;!,  x^  •••  in  the  expansion  of  (1  +  x)^. 

1.  Prove  that  Ci  —  2  C2  +  3 cs 4- (—  l)"-inc„  =  0. 

2.  Prove  that  Co  -  2ci  +  3c2 +  (  -  l)''(n  +  2)c„  =  0. 

3.  Prove  that  co  +  -ci  +  -C2  +  —  -f-  -^—Cn  =  ^"^^  ~  ^' 

2         3  71  +  1  n  +  l 

4.  Prove  that  ^  + 2^+ 3^+ .••  +  n-^  =  -n(n  +  1). 

Co  Ci  C2  C„_i       2 

6.  Provethat^«  +  |+|+"-«  +  ...=-^.     " 
13      6      7  n  +  l 

6.  Prove  that 

Co  +  cix  +  2  Cix"^  +  •••  +  rCrX*"  •••  +  wc„x"  =  1  +  naj(l  +  a;)"-^. 

7.  Prove  that  CoCi+CiC2  +  C2C3+ h  Cn_iCn  =, -.'T      -x.' 

(n  +  l)!(/i  — 1)! 


8.   Prove  that  C0C2+C1C3+C2C4  +  •••  +  c„_2Cn  = 


(n+2)!(n-2)! 


THE  BINOMIAL  THEOREM.  507 

9.   Prove  that  C(?  -  cx^ -^  c^^ h  (-  V)^Cr?  is  equal  to  0  if 

n  ' 
n  be  odd,  and  to  — — ^^ —  if  n  be  even. 

ln\\n\ 

10.  Prove  that 

(ji  —  \)\n\ 

11.  Prove  that 


(2  71-1)! 


(w-l)!(w-l)! 

12.  Prove  that 

13.  Prove  that 

1  Cl         J         C2       ,    /-        l>vn       C**       _  Wj 

X     x+l     x-{-2  ^  x+n     x(x+l)(x-\-2) -.{x+n) 

14.  Show  that,  if  there  be  a  middle  term  in  the  expansion  of 
(1  +  a:)",  its  coefficient  will  be  even. 


THE  BINOMIAL  THEOREM:    ANY   INDEX. 

/    354.   In  Arts.  345,  349,  we  derived  the  series 

1-2  r ! 

as  the  expansion  of  (1  +  a;)**,  but  on  the  hypothesis  that 
n  was  a  positive  integer ;  and  under  this  limitation  the 
series  necessarily  terminates  at  the  (n-\-iy^  term,  for 
the  (n  +  2)"'^  term  and  all  subsequent  terms  contain 
w  —  n  as  a  factor. 

When  n  is  not  a  positive  integer,  the  series  is  endless, 
since  no  one  of  the  factors  n,  n  —  1,  n  —  2,  etc.,  can  in 
that  case  be  zero.  Hence,  to  prove  the  binomial  theorem 
for  a  fractional   or  negative  index  is  a  proposition  in 


508  THE   BINOMIAL  THEOREM. 

infinite  series,  and  any  demonstration  that  may  be  pro- 
posed can  only  be  valid  under  conditions  that  render  the 
infinite  series  convergent.  This  limitation  must  he  pre- 
supposed in  all  the  demonstrations  that  follow. 

f  355.  Before  proceeding  to  give  a  proof  of  the  bino- 
mial theorem  when  the  index  is  fractional  or  negative, 
we  give  some  examples  of  its  application.  In  these 
examples  x  must  be  understood  t6  have  only  such  values 
as  make  the  series  convergent. 

/        Ex.  1.   Expand  (l+cc)-i  by  the  binomial  theorem.    Putn=  — 1 
in  the  formula,  and  we  have 

(1  +  ,)-!  =  1  +(-  1).  +  (-  r>(-  2),.  ^  (-  IX-  2X-3)^3  + ... 

1  '  Z  L  •  Z  '  o 

^(-l)(-2)-(-r)^^... 
rl 
=  1  -  a;  +  a:'-^  -  a:3 +  ...+(- lya?- -f  .... 

I     Ex.  2.  Expand  (1  +  x)-^. 

We  have 
(1  +  ,)-.  =  1 +(_  2).  +  tl21(^,.  +  t.2X-3X^^3  ^ ... 
i.  •  Z  1  •  J  •  o 

+  (-2)C-3)(-4).-C-r-l)^.  _^  _ 
rl 
=  l-2a;-f  3a;2-4a:3  +  ...  4-(_  l)r(,.  +  i)a:'- +  .... 

/     Ex.  3.   Expand  (1  +  x)-K 

(1  +  ,)-3  =  1  +  (_  3).  +  £^|K^  x»  +  £^8K^^iK^.3  + ., 

r\ 

=  J_(l .  2  -  2  .  3ic  +  3  .  4ic2  -  4  .  6a:3  +  ... 

•      +(-iy(r  +  l)(r  +  2)a?- +...}. 


THE  BINOMIAL  THEOREM.  509 


Ex.  4.  Expand  (1  -  x)  K 


(1  - x)-^  =  1  +(_  i)( _ aj)+  (-^X-l)  (_ xy 


1-2 


+  (-i)(-f)(-f)(-:,)3+...  +  (-^)(-|)-(-i-^+l)(_a;W 

-1  I  ^x  I  ^'^x'^  I  ^•^•^x'^  1         I  l-3-5-(2r-l) 
-^^2''^2.4''  +2.4.6'^  +       +      2.4.6...2r      "^^     * 


356.  Euler's  Proof  of  the  Binomial  Theorem.  For  our 
present  purpose  we  shall  employ  the  functional  symbol 
f{n)  to  denote  the  series 

1  +  "^+     lT2    ^+         172^ ^^+-' 

that  is,  the  entire  series  if  n  be  positive  and  integral,  or 
its  limit  [Art.  310]  if  n  be  fractional  or  negative.  In 
this  general  form,  with  n  unrestricted  in  value,  the  series 
is  called  the  binomial  series. 

Let  there  be  three  binomial  series  formed  as  follows : 

/(n)=l+n«+?^i^^  x'  +  ...  (i.), 

/(m)=  l  +  mx  +  V^i^ZzH-^  +  ...        (ii.), 
.     \     ^  .  /      .     \     I   (7n-{-n)(m-{-n—l)   o  ,       ....  ^ 


and  let  x  be  assumed  to  have  only  such  values  as  will 
make  them  absolutely  convergent.  We  may  then  multiply 
(i.)  and  (ii.)  together,  and  their  product  will  be  a  converg- 
ent series  [Art.  324]. 

Now,  if  the  series  on  the  right  of  (i.)  and  (ii.)  be  mul- 
tiplied together,  and  the  product  be  arranged  according 


510  THE  BINOMIAL  THEOREM. 

to  ascending  powers  of  x,  the  result  must  involve  m  and 
n  in  the  same  manner  whatever  their  values  may  be. 
But,  when  m  and  n  are  positive  integers,  we  know  that 
/(m)  is  (1  -f  x)'^  and  that  f{n)  is(l  +  x)  "",  and  the  prod- 
uct of /(m)  and  f{n)  is  therefore  in  this  case  (1  +  a;)"'+% 
which  again,  as  m  +  n  is  a  positive  integer,  is  f{m  -f  n). 
Hence  when  m  and  n  are  positive  integers,  the  product 
/(m)  X  f{n)  is  /(m  +  n)  ;  and  as  the  form  of  the  product 
is  the  same  for  all  values  of  m  and  7i,  it  follows  that 

/(m)  X  f{n)  =  /(m  +  n)  (a), 

for  all  values  of  m  and  7i,  for  which  (i.)  and  (ii.)  are  abso- 
lutely convergent. 

For  the  purposes  of  this  proof  we  need  not  enquire 
what  values  of  x,  m  and  n  will  render  the  series  under 
discussion  absolutely  convergent.  But  see  Art.  323, 
Ex.1. 

By  continued  application  of  (a),  we  have 

J(rn)  X  f{n)  xf{p)  X  -"  =f{m -{-n)  xf(p)  X  — 
=/(m  +^+P  +  ---)- 
Now  let  m  =  n  =  p  =  •  •  •  =  r/s,  where  r  and  s  are  posi- 
tive integers ;  then  taking  s  factors,  we  have 

/(- I  x/|-)X  •••to  s  factors  =/(    -\ — | — tosterms); 
\sj        \sj  \s    s  p 


•■{ 


n\\\  =/w- 


But  since  r  is  a  positive  integer,  /(r)  =  (1  -f-  a?)'. 

.-.  (1 +(.)'=  j/*" 


•••  (i+'»)^=/(; 


THE  BINOMIAL  THEOREM.  511 

This  proves  our  theorem  for  any  positive  fractional 
exponent. 

Next,  assuming  that  the  binomial  theorem  is  true  for 
any  positive  index,  it  can  be  proved  to  be  true  also  for 
any  negative  index. 

For,  from  (a), 

/(-n)x/(n)=/(-n  +  7i)=/(0). 
Hence,  as  /(O)  is  clearly  1,  we  have 

•^^"""^  ""7^  ^  al^r  ''''''^ "" ''  positive, 

=  (l  +  a;)-. 

Thus  (1  4-  «)"'*=/(— n),  which  proves  the  theorem  for 
any  negative  index. 

Ex.  1.  Show  that  the  coeflScients  of  2c«-i  in  the  expansions  of 
(1  —  a:)-"  and  (1  +  x)2«-2  are  equal,  n  being  any  positive  integer. 

The  coefficients  are 
(;-nX-n-l)...(-n-n+2),      .,.,  ^^^        (2n-2)!       ^,,^ 
(w-1)!  ^       ^  (n-l)!(n-l)! 

The  former  =  n(n  +  l)...(2n  -  2) ^„.,  ^        ^2n-2)l 

(n-1)!  (n-l)!(n-l)I 

/       Ex.  2.  Find  -^101  by  the  binomial  theorem. 
V101=  V{100(1  +  jh)}  =  10(1  +  xk)* 

=  10(1  +  .005  -  .0000125  +  .0000000625}  =  10.04987562.... 

Ex.  3.   Find  the  coefficient  of  x^  in  the  expansion  of 
(1  -  2  x  +  4  x2)-2 
according  to  ascending  powers  of  x. 


512  THE   BINOMIAL   THEOREM. 

(1  -  2 X  +  4 x2) -2=  {1  -  2  a;(l  -  2  x)]-^ 

=  1  +  4  x(l  -  2  a:)  +  12  x\l  -  2  x)2  +  32  x^(l  -2xy+"' 
=  1  +  4x  +  4x2  -  16x3  +  .... 

Ex.  4.  Show  that,  if  (1  +  x  +  x2  +  x^)-^  be  expanded  in  a  series 
of  ascending  powers  of  x,  the  coefficients  of  x^  and  x^  will  be  zero. 

(1  +X  +  x2  +  x3)-2=/lj:i^y^=(l-X)2(l  -x4)-2 

=  (l-2x  +  x2)(l+ 2x^  +  3x8  +  4x12  4-...). 

EXAMPLES  XCII. 
1 1.    Expand  (1  —  x)~*  to  5  terms. 
;  2.   Expand  (1  +  2x)~*  to  5  terms. 
'  3.   Expand  (2  —  x)-^  to  6  terms. 

4.  Expand  (1  —  3x)~^  to  5  terms. 

3 

5.  Expand  (1  —  5x)^  to  5  terms. 

Find  the  general  term  in  the  expansion  of  each  of  the  following : 
-    6.    (l-x)-5.        7.   (1-x)-".        8.  (1-x)'^       9.   (l  +  x)2. 
10.  (l+x)"3.  IL   (l-2x)"l  12.   (l  +  3x)"3. 

13.   Find  the  coefficient  of  x^  in  the  expansion  of 


according  to  the  ascending  powers  of  x.  "^  '^  '' 

14.    Expand  (1  —  3x)^  to  5  terms,  and  write  down  the  general 
term  in  its  simplest  form. 

16.   Find  the  coefficients  of  x^  and  x*  in  the  expansion  of 
(1  -2x  +  3x2)-i. 


THE   BINOMIAL   THEOREM.  613 

16.  Find  the  coefl&cients  of  x^  and  x*  in  the  expansion  of 

17.  Find  the  first  negative  term  of  (1  +  ^xy. 

18.  Find  the  first  negative  term  of  (1  +  Sx)'^. 

I       19.   Find  each  of  the  following  to  four  places  of  decimals  by 
means  of  the  binomial  theorem  : 

(i.)   ^110,     (ii.)   ^130,     (iii.)  ^630. 

20.  Find  the  coefficients  of  ofi  and  x^  in  the  expansion  of 

(1  +  X  +  X'  +  X8)». 

21.  Show  that  in  the  expansion  of  (1  +  x  +  x^)-!  in  a  series  of 
ascending  powers  of  x,  the  coefficients  of  x^  and  x*  are  zero. 

22.  Show  that  in  the  expansion  of  (1  +  x  +x2  +  x'-  +  x^  +  x*)"'* 
in  a  series  of  ascending  powers  of  x,  the  coefficients  of  x*  and  x^ 
are  zero. 


2 


614     EXPONENTIAL   AND   LOGARITHMIC   SERIES. 


CHAPTER   XXXV. 

Exponential  and  Logarithmic  Series. 

357.   The  Exponential  Series  is  an  infinite  series  whose 
first  term  is  1  and  whose   {n  +  1)*^  term  is  x^'/n !.     It  is 
absolutely  convergent  for  all  finite  values  of  x  [Art.  3: 
Ex.  2].     Let  its  limit  be  denoted  hj  f{x)  ;  then  the  tw 
exponential  series 

/(x)=i+x+i;+fi+- +!;+•..    (i.) 

may  be  multiplied  together  and  their  product  will  be  a 
convergent  series  [Art.  324]. 

Compare  the  terms  of  this  product  with  the  terms  of 
the  series 

/(a^+2/)=l+(aj+2/)  + ^^±p^+ •- +  ^^±^ 

The  coefficient  of  a;'"^/"  in  f(x)  x  f(y)  is  clearly 
l/{mln\)  and  in  f(x  +  y)  the  term  (x  +  ?/)'"+"/ (m  +  n) ! 
is  the  only  one  in  which  x^y""  can  occur,  and  by  the  bino- 
mial theorem  for  a  positive  integral  index,  the  coefficient 
of  x'^y''  in  {x  +  yy+''/(m  +  n)!  is 

1^  .  1"^,   that  is,  4^. 
(m  +  n) !        mini  mini 


^^ 


p 


EXPONENTIAL   AND   LOGARITHMIC    SERIES.     515 

Hence,  whatever  x  and  y  may  be,  the  coefficients  of 
y,myn  ^j.^  y^^^  ^  J^y^^  ^^^  ^j^  y^^  _j_  ^^  g^j.g  ^j^g  same,  however 

large  m  and  ?i  may  be,  that  is,  whatever  (integral)  values, 
between  1  and  +  oo,  m  and  n  may  have.  It  follows  that 
the  series  (iii.)?  with  its  terms  expanded,  is  identical 
with  the  series  formed  as  the  product  of  the  series  (i.) 
and  (ii.)  ;  hence 

/Wx/(2/)=/(a;  +  2/),  (iv.) 

for  all  values  of  x  and  y. 

By  repeated  application  of  (iv.)  we  obtain 

/W  xf(y)xm  X  ...  =f{x  +  y)  x/(0  X  ... 

=/(a;  +  2/  +  ^  +  -). 

Therefore,  if  z  be  any  positive  integer, 

/(I)  x/(l)  x/(l)  X  to  z  factors 

=/(l  +  1  +  1  +  ...  to  z  terms), 

that  is  \f(l)l'=f(z). 

But  /(l)  =  l  +  l+l  +  l+. ..+  !+... 

—   ^  zl      61  nl 

=  e.  [Art.  305.] 

...  e*=/(0)=l  +  z  +  — +  ...+^  +  ..., 

J\  )         ^        2!  n\ 

if  z  be  a  positive  integer. 

If  z  he  a  positive  fraction,  say  z  =p/q,  where  p  and  q 
are  positive  integers,  then 

f(t\  X  f(^\  X  f(^  xtoq  factors 

\qj     \qj     \qj 


=f(^^+\^\  +  '''^oqi..u..y 


516     EXPONENTIAL  AND   LOGARITHMIC   SERIES. 
•  =  e^,  \;  p  is  Si  positive  integer ; 


Hence  e'=f{z),  for  all  positive  values  of  z  tjotl 
integral  and  fractional. 

If  z  be  negative,  say  z  =  —  z',  then  by  (iv.), 

/(-^')X/(2')=/(0), 
and  y(0)  is  obviously  equal  to  1. 

.*.  /( —  z')  =  — —  =  — ,    '.'  y  is  positive. 
f{z')      e^'         ^      ^ 

Hence,  for  all  rational  values  of  z,  be  they  integral  or 
fractional,  positive  or  negative, 

«'  =  /(.)  =  !+. +  |^+-+J  +  -. 

Should  z  be  irrational,  we  replace  it  by  a  rational 
fraction,  which  may  be  so  chosen  as  to  differ  from  the 
given  irrational  value  by  an  arbitrarily  small  quantity. 
[See  Art.  249.] 

358.  We  may  now  derive  an  infinite  series,  in  ascend- 
ing powers  of  x,  for  the  exponential  f auction  a^. 

Let  A  =  In  a ;  then  a  =  e\  and  a^  =  e'^* ; 

where  A  depends  only  upon  a.  This  theorem,  by  means 
of  which  a""  can  be  expanded  in  a  series  of  ascending 
powers  of  x,  is  sometimes  called  the  exponential  theorem. 


EXPONENTIAL  AND   LOGARITHMIC   SERIES.     517 

359.  The  Logarithmic  Series.     Let  a  =  e^,  so  that 

X  =  In  a. 
Then  a==  =  e^^  =  e^^°«. 

Hence,  from  Art.  357,  we  have 

^x  ^  ^\na  =l-fajlna  +  i  (a;lna)'^  +  —^  (xlnay  +  .... 
Now  put  a  =  l-\-yj  then  we  have 

(l+2/)'==l  +  ajln(l4-2/)  +  |yla;ln(l+.v)r  +  -. 

Now,  provided  y  he  numerically  less  than  unity,  (1  +  y)* 
can  be  expanded  by  the  binomial  theorem;  we  then  have 

^   ,         ,  x(x  —  l)   »  ,  x(x  —  l)(x  —  2)   ,  , 
l-hxy-\-    ^^^  ^y^-\-    ^     -^,2.3 y^^'" 

=  l4-a;ln(l  +  2/)+||{a;ln(H-2/)j=^+-. 

Equate  the  coefficients  of  x  on  the  two  sides  of  the  last 
equation.     [Art.  330.]     We  thus  obtain 

\n{l  +  y)  =  y-yl  +  t-t^.... 

This  is  called  the  logarithmic  senes. 

360.  In  order  to  diminish  the  labour  of  finding  the 
approximate  value  of  the  logarithm  of  any  number,  more 
rapidly  converging  series  are  obtained  from  the  funda- 
mental logarithmic  series. 

Changing  the  sign  of  y  in  the  logarithmic  series 

lna  +  y)  =  y-l  +  t-y^  +  ...,  (i.) 

we  have        ln(l  —  2/)  =  — 2/  — ^  — ^  — ^ ...  (ii.) 

2      3      4 


618     EXPONENTIAL  AND   LOGARITHMIC    SERIES. 
Hence  lni±^  =  ln(l  +  y)-  ln(l  -  y)    [Art.  297,  lY.] 

Put  -  for  i±l,  and  therefore  *^?i^:^  for  y:  then  we 
n         1  —  2/  m  +  n 

have  from  (iii.) 

71  (  m  +  n      3  \m  +  ny      5\m-\-nJ  ) 

361.  We  are  now  able  to  calculate  logarithms  to  base 
e  without  much  labour. 

Put  m  =  2,  n  =  1,  in  formula  (iv.);  then 

ln2  =  2U  +  l.l+l.l  +  .4, 
13     3    33^5    35  /* 

from  which  it  is  easy  to  obtain  the  value 

In  2  =  .693147.... 

Having  found  In  2,  we  have  from  (iv.) 


ln§  = 
2 

-{hi 

.1  +  1. 

53     6 

h- 

•}' 

.-.  In3- 

-ln2  = 

.405465... 

Hence 

ln3  = 

In  2 +  .405465... 

= 

.693147... 

+  .405465...= 

1.098612.... 

Again, 

putting  m  = 

=  5,  w  =  3, 

we  have 

H- 

-IM' 

■h-l 

h- 

. 

= 

=  H^-^8 

"^1280 

f  1  +•■ 

28672 

••} 

= 

=  .510826.. 

.  ; 

EXPONENTIAL   AND   LOGARITHMIC    SERIES.     519 

.-.  In  5  =  1.098612  ...  +  .510826  ... 
=  1.609438  •..  ; 
.-.  In  10  =  ln5  +  ln2 

=  1.609438... +  .693147.. . 
=  2.302585 .... 

By  such  processes  as  these  the  natural  logarithm  of  any  number 
can  be  found  to  any  requisite  degree  of  approximation. 

362.  In  Art.  302  it  was  shown  that  In  10  is  the  recip- 
rocal of  the  modulus  of  the  system  of  logarithms  whose 
base  is  10.     Hence  the  modulus  of  common  logarithms  is 

M=  1/lnlO  =  1/2.302585... 

=  .434294.... 

EXAMPLES  XOIII. 


31^5!      71      91^ 

'•    '  =  ^+2T  +  3l  +  4l+*- 

2        ^31      5!      7! 
5.    (i  +  .).^  =  i+|^  +  3^2^4^^ 

6       1  =  -L_lJ_  +  J_+J__L 

■   2      1.3     3.5      6.7      7.9  * 

7.    ln2  =  -^-f--^+ J-+ J-+.. 
1.2      3.4      5.6      7.8 


520     EXPONENTIAL   AND   LOGARITHMIC   SERIES. 


8.   In2=-  +  — J^ — +       ^ 

2      1.2.3      3.4.5 


X  l2a;+l      3 


1 


6.6.7 
1 


+ 


3(2x+l)3      5(2x+l)5 
10.    ln-^  =  2|— ^— +  i ^ +  1  ^ 


11.   \nx  = 


1_^1    a;2 


(2: 
1    ^1    a:3 


1)3     5(2aj2_i)6 


■■•}• 


1 


+ 


12.  in«+^^    2ax 

a  -  X     a2  4.  a;2     3 


X  +  1      2  (X  +  1)2     3  (x  +  1)» 

1/    2  ax    y  I  1/    2  ax    Y 

13.   -logx  =  -^{^  +  l(^V+...}. 
lnalx+1     3Vx  +  l/  J 


14.    K 


where 
prove  that 


«  =  -i  +  ^\^r3, 


M  +  r  4-  w  =  c*, 

w  +  0)1?  +  w^w?  =  e««>==, 

w  +  w^v  +  wto  =  6"'=*, 
and  that  w^  +  ^3  +  t(;3  —  3  m2;w7  =  1. 

15.   Eind  the  expansions  in  ascending  powers  of  x  for  m,  v, 
and  w  of  example  14. 


LOGARITHMIC   COMPUTATION.  521 


CHAPTER   XXXVI. 

Logarithmic  Computation. 

363.  The  logarithms  used  in  all  theoretical  investi- 
gations are  natural  logarithms;  but  for  the  purpose  of 
making  approximate  numerical  calculations,  for  reasons 
that  will  shortly  appear,  logarithms  to  base  10  are 
always  employed.  On  this  account  logarithms  to  base 
10  are  called  common  logarithms.* 

We  have  shown,  in  the  preceding  chapter,  how  natural 
logarithms,  or  logarithms  to  base  e,  can  be  found ;  and 
having  constructed  a  table  of  natural  logarithms,  the 
logarithms  to  base  10  are  obtained  by  multiplying  the 
former  by  the  modulus  of  the  latter,  that  is,  by  the  con- 
stant factor  ^''log  e  [Art.  291,  VI.],  whose  numerical  value 
has  been  found  to  be 

''loge  =  -^=AS^29-".    [Art.  362.] 
In  10 

common  logarithms. 

364.  In  what  follows  the  logarithms  must  always  be 
supposed  to  be  common  logarithms,  and  the  base  10  need 
not  be  written. 

*  Also  Briggian  logarithms  in  honour  of  Henry  Briggs,  of  Oxford, 
who  first  constructed  tables  of  such  logarithms. 


522  LOGARITHMIC   COMPUTATION. 

If  two  numbers  have  the  same  figures,  and  therefore 
differ  only  in  the  position  of  the  decimal  point,  the  one 
must  be  the  product  of  the  other  *and  some  integral 
power  of  10,  and  hence  from  Art.  297,  III.,  the  logarithms 
of  the  numbers  will  differ  by  an  integer. 

Thus  log 421.5  =  log  4.215  +  log  100  =  2  +  log 4.215. 
Again,  knowing  that  log  3  =  .30103,  we  have 
log. 03  =  log(3  ^  100)  =  .30103  -  2. 

On  account  of  the  above  property,  common  logarithms 
are  always  written  with  the  decimal  part  positive.  Thus 
log  .03  is  not  written  in  the  form  -1.69897,  but  2.30103, 
the  minus  sign  referring  only  to  the  integral  portion  of 
the  logarithm,  and  being  written  above  the  figure  to 
which  it  refers. 

Definition.  When  a  logarithm  is  so  written  that  its 
decimal  part  is  positive,  the  decimal  part  of  the  logarithm 
is  called  the  mantissa,  and  the  integral  part  the  character- 
istic. 

365.  The  characteristic  of  the  logarithm  of  any  number 
can  be  written  down  by  inspection. 

First  let  the  number  be  greater  than  1,  and  let  n  be  the 
number  of  figures  in  its  integral  part ;  then  the  number 
is  clearly  less  than  10"  but  not  less  than  10**-^  The 
logarithm  of  the  number  is  therefore  between  n  and 
71  —  1 ;  thus  the  logarithm  is  equal  to  n  —  1  +  a  decimal. 

Thus  the  characteristic  of  the  logarithm  of  any  number 
greater  than  unity  is  one  less  than  the  number  of  figures  in 
its  integral  part. 

For  example,  235  is  greater  than  10^  but  less  than  10^.  Hence 
log  235  =  2  +  a  decimal,  so  that  the  characteristic  is  2. 


LOGARITHMIC   COMPUTATION.  523 

Next,  let  the  number  be  less  than  1.  Express  the 
number  as  a  decimal,  and  let  n  be  the  number  of  ciphers 
before  its  first  significant  figure.  Then  the  number  is 
less  than  10"**  and  not  less  than  10~"~\ 

Hence,  as  the  decimal  part  of  the   logarithm  ^  must  he 
positive^  the  logarithm  of  the  number  will  be 
—  (n  +  1)  +  a  decimal  fraction, 
the  characteristic  being  —  (n  +1). 

Thus,  if  a  number  less  than  unity  he  expressed  as  a 
decimal,  the  characteristic  of  its  logarithm  is  negative  and 
greater  hy  one  than  the  numher  of  ciphers  hefore  the  first 
significant  figure. 

For  example,  .02  is  greater  than  10-2  i,ut  less  than  lO-i ;  hence 
log  .02  is  —  2  +  a  decimal,  the  characteristic  being  —  2.  Also 
.00042  is  greater  than  10-*  but  less  than  10-3  ;  hence  log  .00042  is 
—  4  +  a  decimal,  the  characteristic  being  —  4. 

366.  Conversely,  if  we  know  the  characteristic  of  the 
logarithm  of  any  number  whose  digits  form  a  certain 
sequence  of  figures,  we  know  where  to  place  the  decimal 
point. 

For  example,  knowing  that  log  1.1467  =  .0594408,  we  know  that 
the  number  whose  logarithm  is  3.0594498  is  1140.7,  and  that  the 
number  whose  logarithm  is  4.0594498  is  .00011467. 

367.  Tables  are  published  which  give  the  logarithms 
of  all  numbers  from  1  to  99999  calculated  to  seven  places 
of  decimals  :  these  are  called  "  seven-figure  "  logarithms. 
For  many  purposes  it  is,  however,  sufficient  to  use  five- 
figure,  or  even  four-figure,  logarithms. 

In  all  Tables  of  logarithms  the  mantissae  only  are  given, 
for,  as  we  have  seen,  the  characteristics  can  always  be 
written  down  by  inspection. 


524  LOGARITHMIC   COMPUTATION. 

In  making  use  of  tables  of  logarithms  we  have,  I.  to 
find  the  logarithm  of  a  given  number,  and  II.  to  find  the 
number  which  has  a  given  logarithm. 

I.  To  find  the  logarithm  of  a  given  number. 

If  the  number  have  no  more  than  five  significant  fig- 
ures, its  logarithm  will  be  given  in  the  tables.  But  if 
the  number  have  more  significant  figures  than  are  given 
in  the  tables,  use  must  be  made  of  the  principle  that 
when  the  difference  of  two  numbers  is  small  compared 
with  either  of  them,  the  difference  of  the  numbers  is 
approximately  proportional  to  the  difference  of  their 
logarithms. 

An  example  will  show  how  the  above  principle,  called 
the  Principal  of  Proportional  Differences,  is  utilized. 

Ex.  1.  To  find  the  logarithm  of  1.14673. 

From  the  tables  we  find  that  the  log  1.1467  =  .0594498,  and  that 
log  1.1468  =  .0594877,  and  the  difference  of  these  logarithms  is 
.0000379.  Now  the  difference  between  1.14673  and  1.1467  is 
f'oths  of  the  difference  between  1.1468  and  1.1467.  Hence  to  find 
log  1.14673  we  must  add  to  log  1.1467  three-tenths  the  difference 
between  log  1.1467  and  1.1468,  that  is  we  must  add 

.0000379  X  -^^  =  .0000113. 

Hence  log  1.14673  =  .0594498  -h  .0000113  =  .0594611. 

II.  To  find  the  number  which  has  a  given  logarithm. 
Ex.  2.   Find  the  number  whose  logarithm  is  2.0594611. 

We  find  from  the  tables  that  log  1.  1467  =  .0594498  and  that 
log  1.1468  =  .0594877,  the  mantissa  of  the  given  logarithm  falling 
between  these  two. 

Now  the  difference  between  .0594498  and  the  given  logarithm 
is  .0000113,  and  the  difference  between  the  logarithms  of  1.1467 


LOGAEITHMIC   COMPUTATION.  525 

and  1.1468  is  .0000379.    Hence,  by  the  principle  of  proportional 
differences,  the  number  whose  logarithm  is  .0594611  is 

•  1.1467  +  l^f  X  .0001  =  1.1467  +  .00003  =  1.14673. 

[N.B.  The  approximation  can  only  be  relied  upon  for  one 
figure.] 

Thus  .0594611  is  log  1.14673,  and  therefore  2.0594611  is 
log  114.673. 

Ex.  3.   Find  ^100,  having  given  log  4.6415  =  .6666584 

and  log  4.6416  =  .6666677. 

log  ^100  =  J  log  100  =  §  =  .6666666. 

Now  '  log  4.6416  =  .6666677, 

and  log  4.6415  =  .6606584. 

Hence  log  ^100  -  log  4.6415  =  .0000082, 

and  log  4.6416  -  log  4.6415  =  .0000093. 

Hence  ^100  =  4.6415  +  ff  of  .0001  =  4.64159. 

COMPOUND    INTEREST  AND  ANNUITIES. 

368.  The  approximate  calculation  of  compound  inter- 
est for  a  long  period,  and  also  of  the  value  of  an  annuity, 
can  be  readily  found  by  means  of  logarithms. 

All  problems  of  this  kind  depend  upon  the  three  fol- 
lowing. [The  student  is  supposed  to  be  acquainted  with 
the  arithmetical  treatment  of  interest  and  present  worth.] 

I.  To  find  the  amount  of  a  given  sum  at  compound 
interest,  in  a  given  number  of  years  and  at  a  given  rate 
per  cent. 

Let  P  denote  the  principal,  n  the  number  of  years,  ^  r 
the  interest  of  ^  1  for  1  year,  and  A  the  required  amount. 

Then  the  interest  of  P  for  1  year  will  be  Pr,  and  there- 
fore the  amount  of  principal  and  interest  at  the  end  of 


526  LOGARITHMIC   COMPUTATION. 

the  first  year  will  be  P(l  +  r).  This  last  sum  is  the 
capital  on  which  interest  is  to  be  paid  for  the  second 
year;  and  therefore  the  amount  at  the  end  of  the  second 
year  will  be  [P(l  +  r)  ]  (1  +  r)  =  P(l  +  ry.  Similarly, 
the  amount  at  the  end  of  3  years  will  be  P(l  +  i-y, 
and  at  the  end  of  n  years  will  be  P(l  +  r)". 

Thus  Az=P(l^ry. 

Oor.  If  the  interest  be  paid  and  capitalized  half-yearly, 
it  can  be  easily  seen  that  the  amount  at  the  end  of  n 

years  will  be  P[lH--]   • 

Ex.  Find  the  amount  of  $100  in  25  years  at  5  per  cent  per 
annum. 

Since  the  interest  on  .$  100  is  $  5,  the  interest  on  $  1  is  $  ^j^. 
Hence  the  amount  of  $1  at  the  end  of  the  first  year  will  be 
f  (1  4-  ^V)'  ^"d  at  the  end  of  25  years  will  be  $(1  +  ^)^. 

Hence  ^A,  the  amount  required,  will  be  f$  100(1  +  ^)^. 

Hence  '     log  ^  =  log  100  +  25  log  |^. 

From  the  tables,  we  find  that 

log21  =  1.3222193,  and  log20=  1.3010300. 
Hence      log^  =  2  +  25(1.3222193  -  1.3010300) 

=  2.5297325. 
Now,  from  the  tables,  we  find      log  338.63  =  2.5297254 
and  log  338. 64  =  2.5297383. 

Hence  log  A  -  log  338. 63  =  .  000007 1 , 

and  log  338.64  -  log  338.63  =  .0000129. 

Hence  A  =  338.63  -f  tV^  of  .01 

=  338.63  +  .005  =  338.635. 
Hence  the  amount  required  is  $338,635. 


LOGAEITHMIC   COMPUTATION.  527 

II.  To  find  the  present  value  of  a  sum  of  money  which 
is  to  he  paid  at  the  end^  of  a  given  time. 

Let  A  be  the  sum  payable  at  the  end  of  n  years,  and 
let  P  be  its  present  worth,  the  interest  on  $  1  being  $  r 
per  annum.  Then  the  amount  of  P  in  w  years  "must  be 
just  equal  to  A. 

Hence,  from  I.,     P=  A{l-\-  r)~". 

Ex.  What  is  the  present  worth  of  $  1000  payable  at  the  end  of 
100  years,  interest  being  at  the  rate  of  5  per  cent  per  annum  ? 

P=  1000(1  +  ^^)-^^  =  1000(f^)-i«>. 

Hence      log  P  =  log  1000  -  100 (log  21  -  log  20) 

=  3  -  100(1.3222193  -  1.3010300) 

=  3 -2.11893  =  .88107. 

Now,  from  the  tables,  log  7.6045  =  .8810707. 
Hence  the  required  present  worth  is  $  7.6045. 

III.  To  find  the  present  value  of  an  annuity  of^A  pay- 
able at  the  end  of  each  of  n  successive  years. 

If  the  interest  on  $  1  be  $  r ;  then  from  II. 

the  present  value  of  the  first  payment  is  ^(1  +  r)"^, 

second     .     .     .     A(l-\-r)-^, 

.     n*^   .     .     .     .     A{l  +  r)'\ 

Hence  the  present  value  of  the  whole  is 
^5(l  +  r)-^  +  (l  +  r)-2+...+(l-fr)-«| 

=  ^(l  +  r)-^;-[^  +  y:  =  jjl-(l  +  r)-j. 

Ex.  Find  the  present  value  of  an  annuity  of  $  100  to  be  paid 
for  30  years,  reckoning  interest  at  4  per  cent. 


528  LOGARITHMIC  COMPUTATION. 

The  interest  on  $  1  is  $  jf  q,  therefore  1  +  r  =  1.04. 

Hence  P  =  ^{1  -  (1.04)-30} 

=  2500(1 -(1.04) -30}. 

Now      log(1.04)-so  =  -  30  log(1.04)  =  -  30  x  .0170333 

=  -.510999  =  1.489001. 

From    the    tables,    log  3.0832  =  .4890017 ;    hence    1.489001    is 
log  .30832. 

Hence  P  =  2500  (1  -  .30832)  =  2500  x  .69168 

=  $  1729.20. 


EXAMPLES  XCIV. 

1.  Write  down  the  numbers  whose  logarithms  are  5.3010300 
and  5.3010300  respectively,  knowing  that  log 2  =  .3010300. 

2.  Write  down  the  numbers  whose  logarithms  are  3.2990931 
and  4.2990931  respectively,  knowing  that  log  1.9911  =  .2990931. 

3.  Having  given  log  46854=4.6707467  and  log  46855 =4. 6707559, 
find  log. 0468546. 

4.  Having  given  log  58961 =4.7705648  and  log  58.962  =  1.7705722, 
find  log. 005896 14. 

6.   Find  -^29,  having  given  log  29  =  1.4623980, 

log  19611  =  4.2924776,  and  log  19612  =  4.2924997. 

6.  Find  ^100,  having  given  log  19307  =  .2857148,  and 

log  19308  =  .2857373. 

7.  Having  given  log  2  =  .3010300,  log  3  =  .4771213, 

log  17187  =  4.2352001  and  log  17188  =  4.2352253, 
find  ^15  to  5  places  of  decimals. 

8.  Having  given  log  3.4277  =  .5350028,  log  32483  =  4.5116561, 
and  log  32484  =  4.5116695,  find  v^034277  to  6  places  of  decimals. 


LOGARITHMIC    COMPUTATION.  o29 

In  solving  the  following  problems  use  the  Table  of  Logarithms, 
p.  670,  and  carry  out  the  results  to  four  places  only. 

9.    Find  the  amount  of  $  1  in  100  years  at  5  per  cent  compound 
interest. 

10.  Show  that  a  sum  of  money  will  be  more  than  doubled  in 
18  years  at  4  per  cent  compound  interest. 

11.  Find  the  amount  of  i^SOO  in  10  years  at  4  per  cent  com- 
pound interest,  the  interest  being  payable  half-yearly. 

12.  What  is  the  present  value  of  ^1000  which  is  to  be  paid 
at  the  end  of  15  years,  reckoning  compound  interest  at  3  per 
cent  ? 

13.  What  is  the  present  value  of  an  annuity  of  $  500  that  ceases 
at  the  end  of  25  years,  interest  being  reckoned  at  6  per  cent  ? 

14.  If  the  average  death-rate  per  annum  in  a  city  be  1^^  per 
cent,  and  the  average  birth-rate  be  2f  per  cent,  and  if  there  be  no 
increase  or  decrease  in  the  population  by  migration,  in  how  many 
years  will  the  population  be  doubled  ? 

15.  The  sum  of  ^  100  was  deposited  in  bank  at  compound  inter- 
est on  the  second  of  January  every  year  during  ten  years,  the  last 
deposit  being  made  in  January,  1880.  Then  on  each  succeeding 
second  of  January  (beginning  in  January,  1881)  during  ten  years, 
$  100  was  withdrawn.  Interest  being  reckoned  at  6  per  cent,  what 
amount  remained  on  deposit  January  1st,  1891  ? 

16.  Twenty  annual  payments  of  $500  each  are  deposited  with 
an  assurance  company  for  the  benefit  of  a  person  to  whom,  be- 
ginning with  the  twentieth  year,  the  entire  amount  paid  in,  to- 
gether with  accruing  interest,  is  to  be  returned  in  40  equal  annual 
payments.  Reckoning  interest  at  5  per  cent,  what  should  be  the 
amount  of  each  annual  payment  ? 

Compute  the  following : 


17.    ^32.4. 

19.    ^99.83. 

21.    </91/^0. 

18.    ^8.934. 
2l 

20.    ^92x^77.. 

22.    </(67/9). 

530  CONTINUED   FRACTIONS. 


CHAPTER   XXXVII. 

Continued  Fkactions. 

369.  An  expression  of  the  form  a-\ 


d 


e  +  etc. 
is  called  a  continued  fraction.     This  is  usually  written  in 
the  more  compact  form 

In  the  general  case  the  numerators  b,  d,  f,  •"  and  the 
denominators  c,  e,  g,  -"  may  be  severally  either  positive 
or  negative ;  but  the  continued  fractions  of  greatest 
importance,  and  those  to  which  the  present  discussion 
will  be  confined,  have  the  form 

^1111 

in  which  b,  c,  d,  e,  -"  are  positive  integers  and  a  is  either 
an  integer  (positive  or  negative),  or  zero. 

The  numbers  a,  b,  c,  d,  •••  are  called  the  first,  second, 
third,  etc.,  partial  quotients  respectively. 

A  continued  fraction  is  said  to  be  terminating  or  non- 
terminating  according  as  the  number  of  partial  quotients 
is  finite  or  infinite. 


CONTINUED   FRACTIONS.  531 

The   following    examples   will    show    how   continued 
fractions  may  be  produced. 

Ex.  1.    Convert  74/26  into  a  continued  fraction. 
Reduced  to  its  lowest  terms  this  fraction  is  37/13.     The  process 
is  then  as  follows : 

E  =  2  +  -^  =  2+         '        =2-,^ ' 

13        ^13/11  i,_J_  1  ,_1_ 

11/2  6  +  i 

Ex.  2.  Convert  (—x^-\-x^-\-l)/(x'^-\-l)  into  a  continued  fraction. 
By  the  ordinary  process  of  division  we  have 

-  CC8  +  X2  +  1  ,    ,     ,  X  ,  ,         1 


X 

Ex.  8.  Convert  y/7  into  a  continued  fraction. 
Since  2  is  the  largest  integer  less  than  y/7,  we  write 

V7  =  2+(V7-2)  =  2  + , 

^^  ^  l/(V7-2)* 

and  we  repeat  this  process  upon  the  successively  occurring  surd 
forms  as  many  times  as  may  be  necessary.  Thus,  segregating  in 
each  case  the  largest  integer  less  than  the  fractional  surd  form,  we 
obtain  in  succession 


1 


:ZL±l  =  i^2ZL^  =  i^ L 


V7-2  3  3  3/(v7-l) 

3       ^V7  +  l_.^      V7-l_^      1 

V7-1  2  2  2/(v/7-l) 

^         =:V7  +  1^1        V7-2^^        1 

V7-1  3  3  3/(V7-2) 

^       =^7-l_2  =  4+V7-2  =  4  4-  ^ 


V7-2      ^  "^  l/(V7-2) 

The  fractional  surds  now  recur  in  the  order  in  which  they  have 
appeared  in  this  cycle,  and  the  continued  fraction  is  therefore  non- 
terminating.     Writing  in  the  successive  partial  quotients  we  have 

-/7=2+  —  —  —  —  ^^ ad  inf. 


532  CONTINUED   FRACTIONS. 

By  reason  of  the  recurrence  of  partial  quotients  in 
continued  fractions  of  this  class,  they  are  called  recurring 
or  periodic  continued  fractions. 

370.  Corresponding  to  every  real  number,  commensu- 
rable or  incommensurable,  there  exists  a  continued  fraction, 
either  terminating  or  non-terminating,  to  which  the  number 
is  equal. 

For,  let  X  be  the  given  number.  Then,  if  a  be  the 
greatest  integer  that  does  not  exceed  x,  we  may  write 

X  =  «i  +  ^,*  (i.) 

where  Xi  is  a  positive  number  not  less  than  1,  and  where 
tti  is  a  positive  or  negative  integer,  or  possibly  zero,  this 
last  case  occurring  whenever  X<1. 

A  repetition  of  this  process  with  Xi  gives 

X,  =  a,-\-^,  (ii.) 

where  a^  is  the  greatest  integer  that  does  not  exceed  X2 
and  X2  >  1 ;  and  n  —  1  such  repetitions  result  in  the  gen- 
eral form 

-X'„_i  =  a„  +  — ,  (iii.) 

*  For  example 

where  J  /  <  1  +  V^  <  4, 

1 


and  l~V5=-2+(3-V5)  =  -2-H^3^^^^^^, 

where  ~2<1-V^<-1- 


CONTINUED    FRACTIONS.  533 

with  the  conditions :  a„  =  a  positive  integer  not  >X„  and 
X„  ^  1,  for  all  integral  values  of  n. 

Substituting  from  (ii.)  in  (i.)  we  now  obtain 

and  replacing  Xg  in  this  result  by  the  expression  for  X^ 
derived  from  (iii.)  by  making  71  =  3,  we  have 

«2+    %-f    Xs 

and  in  general,  by  further  repetitions  of  this  process, 
X  =  a  +J_  JL  _1_...J:.. 

^  ^         ^2  +    ^3  +    «4  +  X„ 

This  is  the  required  continued  fraction. 

In  this  result  X„  may  be  an  integer  for  a  finite  value 
of  n  and  the  continued  fraction  therefore  terminating,  or 
X„  may  never  be  an  integer,  however  large  n  may  be,  in 
which  case  the  continued  fraction  will  be  non-termi- 
nating. 

371.  This  process  of  finding  partial  quotients  will  be 
recognized  as  identical  with  that  of  finding  the  greatest 
common  divisor  in  arithmetic.  Now  when  this  process 
is  applied  to  two  integers  that  have  no  common  divisor 
greater  than  1,  the  last  divisor  is  1,  and  the  last  remain- 
der is  0.     Hence : 

Every  commensurable  number  may  be  converted  into  a 
terminating  continued  fraction. 

On  the  other  hand,  since  every  terminating  continued 
fraction  may  obviously  be  converted  into   an   ordinary 


534  CONTINUED   FRACTIONS. 

fraction  with  commensurable  numerator  and  denomi- 
nator, and  no  such  fraction  can  be  equal  to  a  surd, 
it  follows  that : 

A  surd  cannot   be   equal  to  a   terminating  continued 
fraction. 

372.  The  series  of  terminating  continued  fractions 

«ij  0.1  +  — ,    «!  H ; ,  etc., 

formed,  as  here  indicated,  out  of  one,  two,  three,  etc.,  of 
the  partial  quotients  of 

,111 

^2  +  0^3  +  ^4  + 

are  called  convergents ;  ai  is  the  first  convergent,  ai  +  l/a^ 
is  the  second  convergent,  and  so  on.  Written  as  ordinary 
fractions,  the  first  three  convergents  are  easily  found  to  be 

a^a^  4- 1     aia^z  +  ^3  +  <^i 

The  first  convergent,  but  none  of  the  others,  may  be  zero. 

For  the  determination  of  convergents  of  higher  order 

special  methods,  to  be  explained  presently,  are  employed. 

373.  The  convergents  of  a  continued  fraction  of  the 

form  tti  H •••     (tti,   a^,   a^, -"  all  positive)  are 

^2  +  %  + 
alternately  less  and  greater  than  the  fraction  itself. 

For,  the  first  convergent  is  too  small  because  the  frac- 
tional part •  •  •  is   omitted  :    the  second   convergent 

a,  H — is  too  large  because  in  the  fractional  part  a  por- 


CONTINUED  FRACTIONS.  635 

tionof  the  denominator,  namely, ,  is  omitted;  and 

tbe  third  convergent  is  too  small  because  the  part  omitted 
belongs  again  to  the  numerator.  And  in  general,  in  con- 
structing a  convergent,  the  part  of  the  original  continued 
fraction  omitted  belongs  to  the  numerator  or  denominator 
according  as  the  convergent  is  of  odd  or  even  order. 
Hence  the  theorem. 

374.  To  prove  the  law  of  formation  of  the  successive 
convergents. 

Let  the  continued  fraction  be 

.111 

^2  +  %  +    ^4  + 

The  first  three  convergents  are 

«!    aiCtg  +  l     (aia2  +  l)a3  +  ai 
1  ttg  ajj^a  +  l 

the  third  of  which  may  be  formed  from  the  first  and 
second  by  the  following  rule:  To  obtain  the  numerator 
(or  denominator),  add  the  numerator  (or  denominator) 
of  the  first  convergent  to  the  product  of  the  numerator  (or 
denominator)  of  the  second  by  the  third  partial  quotient. 

Denoting  the  successive  convergents  by  Pi/q^,  Pi/qz, 
Pz/^z-)  "•  Pn/Qni  assume  that  this  law  applies  to  the  forma- 
tion of  the  n^^  convergent. 

Then  the  hypothesis  is 

Now  we  may  pass  from  the  n^^  to  the  {n-\-iy^  con- 
vergent by  changing  a„  into  a„  +  l/a„+i ;  and  hence,  if 
the  n*^  convergent  be 

«nPn-l+Pn-2 


536  CONTINUED   FRACTIONS. 

the  (n  +  1)*^  convergent  will  be 

(g,  +  l/n+l)Pn-l  -\-Pn-2    ^  qn+l(^nPn-l  +^-2)  +  Pn-l 
(a„  +  l/a,+i)5n-l  +  qn-2         «n+l(««gn-l  +  g„-2)  +  ^n-l' 

and  by  the  hypothesis  this  is  equal  to 

f^n+APn  +  Pn-l 

Hence,  if  the  law  be  valid  for  the  formation  of  the 
71*^  convergent,  it  is  also  valid  for  the  formation  of 
the  (n  +  1)*^  convergent.  But  the  third  convergent  is 
formed  in  accordance  with  this  law ;  hence  the  law .  is 
valid  for  the  formation  of  every  convergent.    [Art.  145.] 

375.  Properties  of  Oonvergents.  In  proving  the  proper- 
ties of  convergents  we  consider  the  continued  fraction  to 
be 

,111 

and  denote  the  n^^  convergent  by  p^/Qn- 

I.  If  Pn/<Jn  ^6  the  n'*  convergent  to  a  continued  fraction, 
then 

PrSln-l-  Pn-iqn  =  {-^Y- 

For,  by  the  law  of  formation  of  convergents  just 
proved, 

PrSln-X  -Pn-l(ln  =  (^nPn-l  +  Pn-2)qn-l- Pn-l{<^nqn-l+  Qn-2) 
=  (-  1)  (Pn-lQn-2  -Pn-2gn-l) 
=  (-  l)'(i>n-2g«-3 -Pn-S^n-a) 


=  i-ir-\P2q,-Piq2)- 


CONTINTJED  FRACTIONS.  537 

But  p^i  -  Piq2  =  (aitta  +  1)  -  a^a^  =  1  =  ( -  l)^; 

.*.    Pn^n-l  -Pn-iqn  =  (  "  1)" 

II.  Each  convergent  is  in  Us  lowest  terms. 

For,  any  common  divisor  of  p„  and  q^  will  divide 
PnQn-i  —  Pn-iQni  ^liat  is,  unltj ;  and  hence  1  is  the  only 
such  divisor. 

III.  The  difference  between  two  successive  convergents  is 
a  fraction  whose  numerator  is  unity. 

For  Pn    ^  j>n-l  ^  PnQn-l  ^  Pn-lQn  ^         ^ 

TV.  Any  convergent  is  nearer  in  value  to  the  entire  con- 
tinued fraction  than  the  immediately  preceding  convergent, 
and  therefore  nearer  than  any  preceding  convergent. 

For,  the  continued  fraction  itself  is  obtained  from  the 
{n  -\-iy^  convergent  by  putting  a„+i  H •••  in 

the  place  of  a„^.i ;  and  hence,  if  F  denote  the  continued 
fraction 

Un+X-\- -'APn^-Pn-X  ,     , 

Qn+l  +  ^qn 


Qn  +  Qn-l 


where  X  stands  for ,  a  positive  quantity 

less  than  unity. 
Hence 

p         Pn+l  ^Pnfl  +  \Pn        Pn+1 
Qn+l         Qn+1  +  Ag„  qn+1 

^  HPnQn+l  -Pn+lQn)  ^  (  -  1)"X  , 


538  CONTINUED   FRACTIONS. 

and 

Qn         Qn+l  H-  ^9  Qn 

^  (Pn+ign  -  Pngn+l)  ^  (  -  1)"^^ 

9n{qn+l  +  AgJ  qn(qn+l  +  ^Qn)  ' 

But  X  is  less  than  1  and  is  positive,  and  q^+i  is  greater 
than  g„ ;  hence 

TJ7         i^n+1    ^   TjT        Pn 

V.  The  convergents  of  odd  order  progressively  increase, 
but  are  always  less  than  the  continued  fraction;  and  the 
convergents  of  even  order  progressively  decrease,  but  are 
always  greater  than  the  continued  fraction. 

This  proposition  is  a  direct  consequence  of  IV.  and  of 

Art.  373. 

Note.  —  In  enumerating  the  orders  of  the  convergents  of  the 

continued  fraction  ai  H ,  it  is  customary  to  regard  ai  as  the 

az  + 
convergent  of  first  order,  even  when  it  is  zero. 

VI.  The  difference  between  a  continued  fraction  F  and 
its  n*^  convergent  pj  ^n  ^^  ^^ss  than  1/q^n  ^^^^^  greater  than 
1/2  9  V 

For,  by  IV., 

Ji  fsj   — ~ 


qn    qniqn+i  +  MnY 

where  A.  is  positive  and  less  than  unity.    Hence 


9ng«+i  qn    qniqn+i  +  qnY 

and  since  g„+i  >  g„, 

1     >.      XT  P. 


'•^^^^^1^2? 


n+l 


CONTINUED   FRACTIONS.  539 

Another  form  of  this  test  of  error  is  sometimes  useful. 
Since 

5'n+l  =  ttn+l^n  +  ^n-l 

that  is,  q^qn+i  <  a^+iQ^, 

VII.  Thus,  any  convergent  which  immediately  precedes 
a  large  partial  quotient  is  a  near  approximation  to  the 
value  of  the  continued  fraction. 

Ex.  1.  Assign  a  limit  of  the  error  involved  in  assuming  the 
fourth  convergent  as  an  approximation  to  the  value  of 

J    J i__i i_  JL  J_  J 1 

2+1  +  3+15+2+1  +  3+15+"*' 

The  first  four  convergents  are 

1     3     P3^3  +  1^4 
I*    2'    Qs     2  +  1      3' 

16 

11' 

and  by  VI.  F-—< — - —  =  -^' 


3     £2- 
2'    gs 

£4_3 

.3  +  1 
2  +  1 

.4  +  3 

94       3 

.3  +  2 

1 
15 .  112 

[Art.  374.] 


Thus,  the  error  in  taking  \^  as  the  value  of  the  continued 
fraction  is  less  than  1/1815,  or  .00055  +. 

Ex.  2.    Find  a  series  of  fractions  which  converge  towards  the 
true  value  of  y/2. 

Converting  y/2  into  a  continued  fraction,  we  have 

V2  =  l  + V2-l  =  l+     J,  -,==1+  ^ 


V2  +  1  2  + 


V2  +  1 

1  +  ^J-JL.... 

2+2+2+ 


540  CONTINUED  FRACTIONS. 

and  the  first  six  convergents  of  this  are  ^ 

1      3     2-3  +  1^7     2-7  +  3^17 
l'     2'    2  •  2  +  1      5'     2  .  5  +  2      12'    " 
2-17  +  7^41     2-41  +  17  ^99 
2.12  +  5     29'    2.29+12     70' 
the  last  of  which  differs  from  y/2  by  a  quantity  that  is  less  than 
1/(2  X  702),  or  1/9800. 

376.   A  non-terminating  continued  fraction  of  the  form 

ai H has  a  true  limit,  namely,     _      {Pn/%) • 

In  V.  of  the  preceding  article  it  was  shown  that 

P2n^    ■pi^P2n+l 
Q2n  Q2n+1 

for  all  values  of  n  (P2n/Q2n  and  i?2n+i/^2n+i  representing 
the  convergents  of  even  and  odd  orders  respectively), 
and  in  VI.  it  was  also  shown  that 

and  F-?^<^. 

Q2  n+1         Q  2  n+1 

Adding  the  corresponding  members  of  the  last  two 
inequalities  together,  we  have 

P2n      ^2«+l    .J_   ,   _J_ 

—  ~~  ^^   2      '     2      ' 

92  n         Q2n+1         Q.  2n         9  2  n+1 

Now,  q2n  and  q^n+i  are  the  sums  of  products  of  integers 
whose  number  increases  without  limit  with  the  increase 
of  w,  and  we  may  therefore  make  q^n  and  q^n+i  as  large 
as  we  please;  that  is,  we  may  make  the  difference 
i?2n/^2n  "  i>2  n+i/g2  n+1  as  Small  as  wc  please,  by  sufficiently 
increasing  n. 


CONTINUED  FRACTIONS.  541 

Hence,  P2n/%n  ^^^  P2n+i/Q2n+i  coiiverge  to  a  common 
value  when  n  =  gc,  the  former  through  increasing,  the 
latter  through  decreasing,  values  [Art.  375] ;  and  since 
the  continued  fraction  is  always  intermediate  in  value 
to  any  two  consecutive  convergents  [Art.  373],  ' 
T-j       limit  /       ,      X       limit  /  ,         ^ 

or,  what  is  the  same  thing. 


EXAMPLES  XCV. 
Convert  the  following  numbers  into  continued  fractions : 


1.  |.      3.  Uh      8.   -I        7.   V17.        9.   Va2  +  1. 

2.  |.      4.  .31.       6.   -n.      8.  2V3.      10.  ^^  -  n^  +  n  +  1 

Calculate  the  successive  convergents  of : 

2+  3+  2+  2 

12.  l+^J_J_-JLl. 

2+  3+  4+  5+6 

13.  i  +  A_JL^J_JL 

3+6+7+9+11 

14.  1+-L    1111 


3+1+15+1+3 


111 


15.  If  Pr/Qr  denote  the  r^  convergent  of — , 

a+  6+  a+  5  + 

show  ih2Ltp2n+2  =  P2n  +  ?><Z2n  and  q^+o  =  ap2n  +  (ab  +  l)q2n- 

16.  If  Pr/Qr  denote  the  r*^  convergent  to  any  continued  fraction, 
prove  that 

Pn+l  -  Pn-l  _Pn, 
Qn+l  -  Qn-l       Qn 


542  CONTINUED   FRACTIONS. 

377.  Kecurring  Oontinned  Fractions.  In  Art.  369  a  con- 
tinued fraction  whose  partial  quotients  recur  in  a  definite 
order  presented  itself  as  the  equivalent  of  a  quadratic 
surd.  The  following  examples  show  how  we  may  pass 
from  the  recurring  continued  fraction  to  the  quadratic 
surd. 

Ex.  1.   Convert into  a  quadratic  surd. 

Let  the  fraction  be  denoted  by  x.    Then 

a  +  X 
whence  cc^  _|_  qj^  _  1  _  0. 

The  roots  of  this  quadratic  equation  are 

2  2 

If  a  be  positive,  the  continued  fraction  is  positive  and  the  posi- 
tive value  of  the  radical  must  be  chosen. 

Ex.2.   Convert into   a    quadratic 

surd.  2+1  +  3  +  2+1  +  3  + 

Let  X  denote  the  value  of  the  continued  fraction.    Then 

^  =  _1_J ^ 

2+1+  3+ic' 

and  the  three  convergents  of  this  are 

3  +  x  +  l        4  +  a; 


h  h  and 


9  +  3X  +  2     ]l  +  3ic 


TT                                         *       4  +  a; 
Hence  


ll  +  3x 

or  3  ^2  +  10  X  -  4  =  0, 

and  the  required  surd  is  the  positive   root  of  this  quadratic 
equation ;  that  is, 


CONTINUED    FRACTIONS.  543 


Ex.  3.   Convert  2+  —  -^ -^-^-i--^ -^-^..- into  a 
1+5+2+1+3+2+1+3+ 
quadratic  surd. 

Let  y  denote  the  entire  continued  fraction,  and  x  the  part  which 
is  recurring ;  then 

=  J_J_J_ 

^      2+1+3  +  **' 

and  2/  =  2  + 


5  +  x 

The  value  of  x  was  found  to  be  —  ^  +  ^^^  in  Ex.  2,  and  the 

o 
value  of  y,  obtained  from  the  second  equation,  is 

^        6  +  a; 
or,  when  x  is  replaced  by  its  numerical  value  and  the  denominator 
is  rationalized, 

119  +  V37. 

^  44 

378.  Every  recurring  continued  fraction  is  equal  to  one 
of  the  roots  of  a  quadratic  equation  whose  coefficients  are 
rational. 

Let  y  denote  the  entire  continued  fraction,  and  x  the 
recurring  part ;  then  y  and  x  have  the  respective  forms 
,11  111  ... 

,11             1         11  r'\ 

x  =  ai-\ >  (ii.) 

in  which  all  the  letters  are  integers,  and  all,  except 
possibly  a,  are  positive. 

lip/q  and  p^ /q'  be  the  convergents  to  (i.)  correspond- 
ing to  the  partial  quotients  h  and  Tc  respectively,  then 

y=—, —y  (ill.) 

^       q'x-\-q  ^      ^ 


(iy.) 


644  CONTINUED  FRACTIONS. 

whence,  by  solving  for  x, 

p-qy  . 

X , 

q'y  -p' 

and  if  r/s  and  r'/s'  be  the  convergents  to  (ii.)  corre- 
sponding to  the  partial  quotients  «„„i  and  a^  respectively, 
then 

X  =   ,    ^  ■•  (v.) 

s'x+s  ^    ^ 

Eeplacing  x  in  (v.)  by  its  value  in  terms  of  y  in  (iv.), 
we  obtain 

P  —  qy      r'(p  —  qy)+r(q'y-p') 


q'y  -p'      s'{p  -  qy)  +  s{q^y  -p') 
a  quadratic  equation  in  y. 

Cleared  of  fractions,  equation  (v.)  is 
s'y? -\-{s  —  r')x  —  r  =  (}'j 

and,  since  {s  —  r'Y  —  4(—  rs')  is  positive  and  —  r  is  neg- 
ative, its  roots  are  real,  one  being  positive,  the  other 
negative.  Hence  the  values  of  y,  as  shown  by  equation 
(iii.),  are  both  real. 

Ex.  1.  Convert       2-ir—  ^  —  —"- 
^1+2+1+2  + 

and  _i+A_J_JLiJ.... 

^3+  1+  24-  1+  2+ 

into  quadratic  surds,  j 

Let       ,  0^  =  2+^  A..., 

and  y  =  -^-^^T^~-"" 

3+  1+  2  + 

Then  x  =  2  +  -l-l,   ^^  ==^  i  +  i  J_  1; 

1+  X  3+  1+  a; 


CONTINUED  FRACTIONS, 


from  which,  by  writing  down  the  successive  convergents,  we  derive 

*=- 

whence,  by  solving  for  x, 

a;2  _  2  ic  -  2  =  0, 
and,  by  solving  for  y, 

yi  -2y-2  =  0. 

Thus  X  and  ?/  are  the  roots  of  the  quadratic  equation 

«2_2«-2  =  0, 

and  since  x  must  be  positive  and  y  negative, 

. •.  X  =  1  +  VS,  y  =  l-y/3. 

Ex.  2.  Express  the  positive  root  of  x^  —  2  a;  —  a^  =  0  as  a  con- 
tinued fraction,  a  being  a  positive  integer. 

If  a,  — /3  denote  the  roots  of  this  equation,  a  its  positive,  — /3  its 
negative  root,  then 

x^-2x-  a2=(x  -  o)(a;  4-  /3), 

and  this  identity  makes  it  at  once  clear  that  x^  —  2x  —  a^  will  be 
negative  for  all  positive  values  of  x  less  than  a.  Hence  the  largest 
integer  less  than  a  is  the  largest  integral  value  of  x  that  will  make 
a;2  —  2  X  —  a^  negative.  This  value  is  easily  seen  to  be  1  +  a,  and 
we  therefore  write 

a  =  l  +  a  +  a  —  1—  a. 

But  a2  -  2  a  -  a2  =  0,  and  therefore 

a2  _  2  a  +  1  -  a2  =  (a  -  1  -  a)  (a  -  1  +  a)  =  1, 

whence  o  =  1  +  a  +  - — ~ 

a  ~1  +  a 
Then 

a-l  +  a  =  2a  + } , 

a  -  1  +  a 
and  therefore 

2a  + ^- 2a+2a  + 

a  —  l  +  a 
2m 


546  CONTINUED   FRACTIONS. 

In  the  larger  text-books  of  algebra  it  is  shown  that  any- 
quadratic  surd  can  be  developed  into  a  recurring  con- 
tinued fraction.     [See  Treatise  07i  Algebra,  Art.  367.] 


EXAMPLES  XCVI. 
Convert  the  following  surds  into  continued  fractions  : 

1.   3  +  V5.  4.   1  -  v/7.  7.   ^V'^'^^. 

V2  -  1 


2.  3-V5.  5.    i(v3-l).         8.   2  +  \/l  +  a2. 

3.  1+V7.  6.   2+-L.  9.   2-VrT^2. 

Convert  the  following  continued  fractions  into  surds : 

10.  *_L  J_  J_  J_  .... 

2+  3+  2+  3  + 
jj     *  1       1        1       1        1        1 


2+2+1+2+2+1  + 

12.  3  +  *-l- J-  JL.... 

5+  5+  5  + 

13.  l  +  J_*A.^^.... 

2+    1+   1+   1  + 

14.  l+A_*_^A_^JL.. 

2+    2+4+2+4  + 

15.  %_!+       1  1 


a+l+  a-l+  a  +  l+  a— 1  + 

Express  the  positive  root  of  each  of  the  following  equations  as  a 
continued  fraction : 

16.  0^2 -2a;- 1  =0.  18.  a;2-2a;-9  =  0. 

17.  a;2  _  2 ic  -  2  =  0.  \Q,   x^  -  4:X  +  ^  -  a^  =  0. 

*  The  asterisk  indicates  the  beginning  of  the  recurring  part. 


CONTINUED   FRACTIONS.  547 

20.  Prove  that 

\a+b+c-\-d+      l\        b+  c+  d-i-      J 

21.  Show  that  a is  a  root  of  the  quadratic 

a  —  a  —  a  — 
equation  x^  —  ax+  1  =0. 

22.  Show  that  the  negative  root  of  the  equation  x^—2x  —  a^=0j 
expressed  as  a  continued  fraction,  (a  =  a  positive  integer)  is 

1  1  1 


a  + 


-2a+   -2a+  — 2a  + 


23.   Express  both  roots  of  the  equation  x^  —  ax  —  a%^  =  0  in  the 
form  of  continued  fractions,  a  and  h  being  positive  integers. 


548  SCALES   OF  NOTATION, 


CHAPTER  XXXVIII. 
Scales  of  Notation. 

379.  In  arithmetic,  any  number  whatever  is  repre- 
sented by  one  or  more  of  the  symbols  0,  1,2,  3, 4,  5,  6,  7, 
8,  9,  called  figures  or  digits,  by  means  of  the  convention 
that  every  figure  placed  to  the  left  of  another  represents 
ten  times  as  much  as  if  it  were  in  the  place  of  that  other. 
The  cipher  0,  which  stands  for  nothing,  is  necessary 
because  one  or  more  of  the  denominations  units,  tens, 
hundreds,  etc.,  may  be  wanting. 

The  above  mode  of  representing  numbers  is  called  the 
common  scale  of  notation,  and  10  is  called  the  radix  or 
base. 

380.  Instead  of  10,  any  other  number  might  be  used 
as  the  base  of  a  system  of  numeration,  that  is,  of  a  sys- 
tem by  which  numbers  are  named  according  to  some 
definite  plan,  and  of  the  corresponding  scale  of  notation, 
that  is,  of  a  system  by  which  numbers  are  represented 
by  a  few  signs  according  to  some  definite  plan. 

Thus  4235  is  said  to  be  written  in  the  scale  of  seven,  if 
3  stands  for  3x7,  2  for  2x7x7,  and  4  for  4x7x7x7, 
so  that  every  figure  placed  to  the  left  of  another  repre- 
sents seven  times  as  much  as  if  it  were  in  the  place  of 
that  other. 


SCALES   OF   NOTATION.  549 

In  general,  any  number  N  is  expressed  in  the  scale  of 
r  when  it  is  written  in  the  form  '"d^d^d^dQ,  where  each  of 
the  digits  df^did^d^'"  is  zero,  or  a  positive  integer  less 
than  r,  and  where  d^  stands  for  c^o  units,  d^  stands  for 
di  X  r,  dz  stands  for  c^a  X  r  x  r,  and  so  on.     Thus 

]Sr=do-^dir-\-d2r^-\-'-'. 

381.  Any  positive  integer  can  be  expressed  in  any  scale 
of  notation. 

Let  N  be  the  number,  and  let  r  be  the  radix  of  the 
required  scale. 

Divide  N  by  r,  and  let  Qi  be  the  quotient  and  d^  the 
remainder. 

Then  N=  d,-\-rx  Q,. 

Now  divide  Qi  by  r,  and  let  Qa  be  the  quotient  and  di 
the  remainder. 

Then  Qi  =  d^-\-r  x  Q^]   .'.  N=  d^  +  rd^  +  r'Q^. 

By  proceeding  in  this  way  we  must  sooner  or  later 
come  to  a  quotient  which  is  less  than  r ;  let  this  be  after 
n  divisions  by  r.  The  process  is  now  complete,  and  we 
have 

N=^  do  +  d,r  +  ^27-2  +  ...  +  d,r-, 

so  that  the  number  would  in  the  scale  of  r  be  written 

d„'"dsd^ido. 

Each  of  the  digits  do,  dy,  c?2,  •  •  •  is  a  positive  integer  less 
than  r,  and  any  one  or  more  of  them,  except  the  last,  d„, 
may  be  zero. 


550 


SCALES   OF  NOTATION. 


Ex.  1.   Express  1062  in  the  scale  of  7. 

The  quotients  and  remainders  of  the  successive  divisions  by  7 
are  as  under : 


1062 
151  remainder  5  =  dQ 
.  .  .  4  =  c?i 
.  .  ,  0  =  do 


21 


Thus,  1062  when  expressed  in  the  scale  of  7  is  3045. 

Ex.  2.   Change  2645  from  the  scale  of  8  to  the  scale  of  10. 

Since  2645  =  2x83  +  6x82 +  4x8  +  5 

=  {(2x8  +  6)8  +  4)8  +  5, 

the  required  result  may  be  obtained  as  follows :  Multiply  2  by  8, 
and  add  6 ;  multiply  this  result  by  8,  and  add  4 ;  then  multiply 
again  by  8,  and  add  5.  This  process  is  clearly  applicable  in  all 
cases. 

Ex.  3.   Express  2156  in  the  scale  of  5.  Ans.  32,111. 

Ex.  4.   Express  34239  in  the  scale  of  11. 

Ans.  237^7,  where  t  is  put  for  10. 

Ex.  5.    Express  \^  as  a  fraction  in  the  scale  of  4.  Ans.  j^. 

Ex.  6.    Change  31426  from  the  scale  of  8  to  the  scale  of  4. 

Ans.  3030112. 

We  may  first  express  31426  in  the  scale  of  10,  as  in  Ex.  1,  and 
reduce  the  result  to  the  scale  of  4,  as  in  Ex.  2.  The  result  may, 
however,  be  obtained  by  one  process,  as  under : 

31426 


4 

6305  remainder  2 

4 
4 

1461 1 

314 1 

4 

63 0 

4 

14 3 

3 0 

Thus,  the  number  required  is  3030112. 


SCALES   OF  NOTATION.  551 

Explanation.  We  first  divide  3  eights  plus  1  by  four,  giving 
quotient  6,  and  remainder  1 ;  we  then  divide  1  eight  plus  4  by 
four,  giving  quotient  3,  and  remainder  0  ;  and  so  on. 

382.  It  would  be  a  good  exercise  to  perform  all  the 
ordinary  rules  of  arithmetic  with  numbers  expressed  in 
various  scales. 

Ex.  1.    Add  2345,  6127,  and  1503.     [Scale  8.] 

Ex.  2.   Subtract  3154  from  4021.     [Scale  6.] 

Ex.  3.   Multiply  234  by  466.    [Scale  7.] 

Ex.4.   Divide  22326 by  315.     [Scale  8.] 

Answers.    12177,  423,  150663,  66. 

383.  Eadix  Tractions.  '  Eadix  fractions  in  any  scale 
correspond  to  decimal  fractions  in  the  ordinary  scale ; 
thus  .abc  •••,  in  the  scale  of  r,  stands  for 

To  show  that  any  given  fraction  may  he  expressed  by  a 
series  of  radix  fractions  in  any  proposed  scale. 

Let  F  be  the  given  fraction;  and  suppose  that  when 
expressed  by  radix  fractions  in  the  scale  of  r,  we  have 

i^=.a6c--.,  that  is  i^=- +  -,  +  -,+  •••, 
r      r^      r^ 

where  each  of  a,  b,  c,  -"  is  zero  or  a  positive  integer  less 
than  r. 

Multiply  by  r ;  then 

r      r- 


552  SCALES   OF   NOTATION. 

Hence   a   must  be   equal   to   the   integral  part,   and 

b      c 

-  +  --  +  •  •  •  must  be  eq  ual  to  the  fractional  part  of  F  xr. 
r      r- 

Let  jp\  be  the  fractional  part  of  Fr ;  then 

r      IT 
Multiply  by  r  again,  then  as  before  h  must  be  equal  to 
the  integral  part  of  F^  x  r. 

Thus  a,  6,  c,  •••  can  be  found  in  succession. 

Ex.  1.  Express  y^^^  by  a  series  of  radix  fractions  in  the  scale 
''^^'         ^\x6  =  3  +  tV;  TVx6=:0  +  i;    1x6  =  2. 
Hence  -302  is  the  required  result. 

Ex.  2.    Change  431.45  from  the  scale  of  10  to  the  scale  of  4. 
The  integral  and  fractional  parts  must  he  done  by  separate 
processes. 

4 

4 

•     4 

4 


431 

.45 

107  ....  3 
26  ....  3 

4 

1.80 

6  ....  2 
1  ....  2 

4 
3.2 

red  result  is  12233.130. 

4 

0.8 

384.  Theorem.  Any  number  expressed  in  the  scale  of  r 
is  divisible  by  r  —  1,  if  the  sum  of  its  digits  is  divisible 
by  r  —  1. 

Let  J^  be  the  number,  S  the  sum  of  the  digits,  and  let 
the  digits  be  do,  di,  d^,  etc. 

Then  N=  do  +  d^r  +  d^r^ -] h  dj-^ 

and  S  =dQ-\-di  +  cZg    +  •  •  •  +  (^n- 

Hence  JSf-S=d,{r-l)+d^Cr''-l)  +  "'-^d„(r''-l), 


SCALES   OF   NOTATION.  553 

Now  each  of  the  terms  on  the  right  is  divisible  by 
r  -  1.     [Art.  147.] 

Hence  iV—  ^  is  divisible  hj  r  —  1,  and  therefore  when 
S  is  divisible  by  r  —  1,  so  also  is  JH. 

As  a  particular  case  of  the  above,  any  number  expressed 
in  the  ordinary  scale  is  divisible  by  9  when  the  sum  of 
its  digits  is  divisible  by  9. 

385.  Theorem.  Any  number  expressed  in  the  scale  of  r 
is  divisible  by  r  -{-  1  when  the  difference  between  the  sum  of 
the  odd  digits  and  the  sum  of  the  even  digits  is  divisible  by 
r  +  l. 

Let  N=  do  +  d^r  +  d^r^  +  d^r^ -\ , 

and  D  =  dQ  —  di-\-do  —  ds-\ . 

Then  N-D=di(r  +  1)  +  d^ii^  -  1)  +  ^3(7^  -j-  1)  +  •••. 

Now  each  of  the  terms  on  the  right  is  divisible  by 
r  +  1.    [Art.  147.] 

Hence,  JV—  D  is  always  divisible  by  r  +  1,  and,  there- 
fore, when  D  is  divisible  by  r  -f  1,  so  also  is  N. 

As  a  particular  case,  any  number  expressed  in  the 
ordinary  scale  is  divisible  by  11  when  the  difference 
between  the  sum  of  the  odd  and  the  sum  of  the  even 
digits  is  divisible  by  11. 

EXAMPLES    XCVII. 

1.  Express  2156,  7213,  and  192457  in  the  scale  of  6. 

2.  Change  the  following  numbers  from  the  scale  of  7  to  the 
ordinary  scale :  2135,  4210,  30012,  123456. 

3.  Change  8152  and  23678  from  the  scale  of  9  to  the  scale  of  12. 

4.  Express  23.42  and  123.45  in  the  scale  of  5. 

20 


554  MISCELLANEOUS   EXAMPLES   VII. 

5.  Multiply  2.31  by  1.25,  the  numbers  being  expressed  in  the 
scale  of  6. 

6.  Show  that  the  numbers  represented  in  any  scale  by  121, 
12321,  and  1234321  are  perfect  squares. 

7.  Find  the  values  of  a  and  6  in  order  that  215al463  may  be 
divisible  by  9  and  by  11. 

8.  Find  the  values  of  a  and  h  in  order  that  516a72456  may  be 
divisible  by  99. 

9.  Find  the  scale  in  which  314  is  represented  by  626. 

10.  Find  a  number  of  two   digits  in  the  scale  of  5  which  is 
doubled  by  reversing  its  digits. 

11.  Find  a  number  of  two  digits  in  the  scale  of  8  which  is 
doubled  by  reversing  its  digits. 

12.  Find  a  number  of  two  digits  in  the  scale  of  7  which  is 
trebled  by  reversing  the  digits. 


MISCELLANEOUS  EXAMPLES  VII. 

1.  Solve  the  equation  x-^^a  =  6  +  ^/(ax^  4-  c). 

2.  Show  that  (x -h  y  +  zy^+^  -  x'^"+i  -  y^»+^  -  z^^+^  is  divisi- 
ble by  (y  -^  z)  (z  +  x)  (x  -\-  y) ,  and  find  the  quotient  when  n  =  1, 
and  when  n  =  2. 

3.  Show  that     (x  +  yy  -x^  -y^  =  S  xy{x  +  y), 

and  that     (x  -\- yY  —  x^  —  y^  =  b  xy{x  +  y)  (oj^  ■\- xy  -\-  y^). 

4.  Simplify  «^ + ^- +  ^' 


{a-b){a-c)      (6-c)(6  — a)      {c—a)(c-b) 
6.   Simplify  ^ + ^« +  «^ 


(a— 6)(a  — c)      (b  —  c){b-a)      {c—a)(c—b) 
6.   Simplify  ^ + i- -f  ^ 


aCa-&)(a— c)      b{b—c)Q)  —  a)     c(c— a)(c-6) 


MISCELLANEOUS   EXAMPLES  VII.  556 


7.    Simplify 

1 


a2(a-6)(a-c)      b\b-c)(ib-a)     c\c-a){c-h) 

8.  Simplify  ^±^ + £±^ + 9i+l 

(a-&)(a-c)      (6-c)(6-a)      {c-d){c-h) 

9.  Simplify  ^±^ + ^±^ + «  +  ^ 

a(a-6)(a-c)      6(6-c)(6-a)      c{c-a){c-h) 

10.   Show  that  "^  ■  ^ 


+ 


(a-6)(a-c)(a;  +  a)      (6_c)(6-a)(x+6) 
1  1 


(c-a)(c-6)(a;  +  c)      (x  +  a)(x  +  6)(x  +  c) 


11.  Show  that  ^ +  ^ 


+ 


(a-6)(a-c)(x  +  a)      (6  -  c)(6  -  a)(x  + 6) 

c  —X 


12.  Show  that 


(c-a)(c-6)(x  +  c)      (x+a)(x+ 6)(x  +  c) 
a2  ,  62 


(a-6)(a-c)(x  + a)      (6  -  c)(6  -  a)(x  +  6) 


{c-a){c-h)(x-\-c)      (x+ a)(x  +  6)(x-t-c) 

13.  Simplify  (6-^-c^)«+(c^-a^)«  +  K-6-^)«. 

(6-c)8+(c-a)8  +  (a-6)» 

14.  Prove  the  following : 
(i.)a  +  6-    «'     ■      *' 


a  —  6      6  —  a 
(ii.)a+6+c=- «^ +  ^'  •  ^' 


aii.)  a  +  6  +  c+c?  = 


{a-h){a-c)      (b-c)(b-a)      {c-a){c-b) 


{a-b){a-c)(a-d)      (b-c){b-d)ib-a) 


+  .       ,..^\.      ,.+  ^^ 


(c-d)(c-a)(c-6)      (d-a)(d-6)(d-c) 


556  MISCELLANEOUS   EXAMPLES   VH. 

a3  ,  68 


15.   Show  that 


(a-6)(a-c)(a-d)      (b  -  c)(b  -  d)Cb  -  a) 
c3  ,  #  ^^ 


(c-£Z)(c-o)(c-6)      (c?-a)((?-&)(<Z-c) 
16.  Show  that  ^^^  ■  ^^« 


(a-6)(«-c)(a-d)      (&- c)(6-d)(6-a) 
(^«6  _l abc _  J 


(c-d)(c-a)(c-6)      ((?-a)(d-6)(d-c) 

17.  Show  that  ^ (x.- ay  + ^ (x-b)^ 

+  7 ^7 -(X-<))2  =  X2. 

(c  —  a)  (c  —  6) 

18.  Prove  that  (b  +  c)2  +  (c  +  ay  +  (a  +  6)^  -  (c  +  a)  (a  +  6) 
-  (a  +  6)(&  +  c)  -  (6  +  c)(c  -f  a)  =  a2  +  62  +  c2  -  6c  -  ca  -  a5. 

19.  Show  that  (6  +  c)3+(c+a)3+(«+6)^-3(6  +  c)(c  +  a)(a  +  6) 

=  2(a3  +  63  +  c3-3a6c). 

20.  Show  that  (b  +  c  -  ay +  (c  +  a  -  by +(a  +  b  -  cy 
-3(6  +  c  -a)(c  +  a-6)(a+  6-c)  =  4(a3  +  63  +  c3-3a6c). 

21.  Show  that  (ic2  +  2?/0)3 +  (y2  +  2  2x)3 +  (^2  +  2xi/)3 

-  3(x2  +  2  yz)  (y^  +  2  zx) (z^  +  2xy)  =  (x^  -\- y^  +  z^  -  S  xyzy. 

22.  Show  that,  if  cc2  _^  ^2  _|_  ^2  _  (^y^  -{-  zx  +  xy),  then  x  —  y  —  z. 

23.  Prove  the  following  : 

(i.)  if  2(a2  +  62)  =  (a  +  6)2,  then  a  =  6  ; 
(u.)  if  3(a2  4.  62  +  c2)  =  (a  +  6  +  c)2,  then  a  =  6  =  c ; 
(iii.)  if  4(a2  +  62  +  c2  +  cZ^)  =  (a  +  6  +  c  +  (Z)2, 
then  a  =  h  =  c  =  d\ 

and     (iv.)  if  n{a'^  +  62  +  c2  +  ...)  =  («  +  6  +  c  +  -O^, 
then  a  =  6  =  c  =  •••, 

w  being  the  number  of  the  letters. 

24.  Prove  that,  if  the  sum  of  two  positive  quantities  be  given, 
their  product  is  greatest  when  they  are  equal  to  one  another. 


MISCELLANEOUS   EXAMPLES   Vn.  557 

25.  Prove  that,  if  the  product  of  two  positive  quantities  be  given, 
their  sum  is  least  when  they  are  equal  to  one  another. 

26.  Show  that  the  least  value  of  x  +  -  is  2,  the  least  value  of 

X 

4  9 

X  +  -  is  4,  and  the  least  value  of  x  +  -  is  6,  x  being  real  and 

X  X  * 

positive. 

27.  Show  that  ^^—^ — ^  "*"     cannot  be  greater  than  7  nor  less 
than  I,  for  real  values  of  x. 

28.  If  the  sum  of  a  given  number  of  positive  quantities  is  fixed, 
their  continued  product  will  be  greatest  when  they  are  all  equal. 

29.  If  the  continued  product  of  a  given  number  of  positive  quan- 
tities is  fixed,  their  sum  will  be  least  when  they  are  all  equal. 

30.  Prove  that,  if  ^^  ~  ^  =  ^^  ~  ^  ,  then  will  each  fraction  be 

y  +  z         0  +  X 

equal  to  ^V  ~  ^  ^  and  also  equal  to  x-\-  y  -\-  z. 
x  +  y 

31.  Prove  that,  if  a,  6,  c,  d  be  all  real  and  if 

(a  +  by  +  (6  +  c)2  +  (c  +  dy  =  4(a&  -hbc  +  cd), 
then  a  =  b  =  c  =  d. 

32.  If       «^-^^      =       ^^  -  ^^     ,  then  will  a  =  6,  or  c  =  tf, 

a-b  -  c  +  d     a  —  b  —  d-{-c 

or  a  +  b  =  c  +  d. 

33.  Show  that,  if  x,  y,  z  be  determined  by  the  equations  : 

(a  -  a)2x  +  (a  -  pyy  +  (a  -  yyz  =  (a  -  5)2, 
(6  -  a)2x  +  (6  -  /3)2?/  +  (6  -  7)2^  =  (6  -  S)^, 
(c  _  a)2x  +(c  -  /3)2?/  +(c  -  7)^0  =(c  -  5)2; 

then  will    (d  -  a)2x  +(d-  /3)2y  +  (c?  -  7)2^  =(d-  5)2, 

where  cZ  has  any  value  whatever. 

34.  If  the  equation  — ^—  +  — ^  =  -^—  +  — ^  have  a  pair  of 

x+a     x+6     x+c     x+d 

equal  roots,  then  either  one  of  the  quantities  a  or  6  is  equal  to  one 


558  MISCELLANEOUS   EXAMPLES   VII. 

of  the  quantities  c  or  d^  or  else  -  +  i  =  -  +  -.     Prove  also  that  the 

a     h     c     d 

roots  are  then  -«,-«,  0  ;  -  6,  -  6,  0  ;  or  0,  0,  -  ^^  • 

«  +  6 

35.  If  2(ic2  +  r,^i2  _  a:a;')  (?/2  +  yn  _  ^y')  =  ^^2^2  +  ^/2y/2^  then  will 
jc  =  oj'  and  y  —  y' . 

36.  Show  that 

{y  -\-  z)(z  +  X) (x  +  y) -\-xyz  =  xyz(x  -\- y  ■\- z)l-  + -^  -  )- 

37.  If  a,  &,  c,  cc  are  all  real  quantities,  and 

{cfl  +  62)x2  -  2  6(a  +  c)a;  +  62  _^  c^  =  0, 
then  a,  6,  c  are  in  geometrical  progression  and  x  is  their  common 
ratio. 

1  -  a;2* 


38.   Show  that  (1  +  x)  (1  +  x^)  (1  +  x^)—  to  n  factors 


1-x 

39.  If  s  =  a  4-  &  +  c,  prove  that 

(as  +  he)  {hs  +  ca)  (cs  +  ah)  =  (6  +  c)2(c  +  ay  {a  +  5)2. 

40.  If  a  +  6  +  c  +  d  =  0  ;  then  will 

(a3  +  53  _|.  c3  +  #)2  =  9(6cd  +  c(?a  +  dah  4-  a&c)2 

=  9(&c  —  ad){ca  —  hd){ah  —  cd) 
=  9(6  +  cy\c  +  a)2(a  +  6)2. 

41.  Show   that    a(a  +  c?)(a  +  2(Z)(a  +  3d)  +  d^    is   a   perfect 
square. 

42.  Prove  (by  the  method  of  indeterminate  coefficients)  that 
the  necessary  and  sufficient  condition  in  order  that 

ax'^  +  2hxy  +  by^  +  2gx  +  2fy-\-c 
may  be  resolved  into  two  factors,  rational  and  of  the  first  degree 
in  X  and  y,  is 

abc  -\-2fgh-  ap  -  hg^  -  ch'^  =  0. 

,43.   Find  (by  the  method  of  indeterminate  coefficients)  the  cube 
root  of  each  of  the  following  expressions  : 
(i.)  x^  -  24  x^y  +  192  xy^  -  512  yK 
(ii.)  x6  -  9 x5  +  33 X*  -  63 x^  +  66 x2  -  36x  +  8. 
(iii.)  8x6  -  36 x&  +  102x4  -  171  x^  +  204x2  _  144x  +  64. 


MISCELLANEOUS   EXAMPLES   VII.  559 

44.  Find  the  limit  of  («  —  a;)^  -  1  ^^^^  x  =  0. 

X 

45.  Find  the  limit  ot  {  x V~^  when  x  =  1. 


i-D- 


46.  Prove    that  ^^^^^^  (1  +  x  +  2  !  x2  +  3  !  a^  +  -  +  n  !a?»)=  oo 
for  all  finite  values  of  x. 

47.  Expand  ^-^  +  ^f-\-^  +  ^  into  an  infinite  series. 

(1  +  x)%l  -  x)8 

48.  Expand  (1  +  ac)'"^  and  (1  +  ic)-'»/2  into  infinite  series. 

49.  Expand  ^e'' +  e-^)  and  ^(e*  —  e-')  into  infinite  series. 

50.  Resolve 

1  4-  ai  +  (1  +  ai)«2  +  (1  +  ai)(l  +  a2)a3  +  - 

+(1  +  ai)(l  +  as)  -  (1  +  an-i)an 
into  n  binomial  factors. 

51.  Find  the  sum  of  the  first  n  4-  1  terms  of  the  series 

1  I   ^   I  (gi  +  ^)^  I  (ai  +  g)  (a2  +  x)x 

.  (ai  +  a;)(a2  +  a;)(a8  +  x)x  ^ 
aia2(Xsci4 

52.  Show  that  the  sum  of  the  first  n  +  1  terms  of 

J  ■  m  .  m(jm  +  1)  .  m{m  +  l)(m  +  2) 
"^  1  ■*■        2 1  3 ! 

m(m  +  l)(m  +  2)(m  +  3)        ^ 
4! 
•g  (m  +  l)(m  +  2)(m  +  3)  ■.■  {m^-n-  1). 

n! 

63.   Show  that  the  sum  of  the  first  n  +  1  terms  of 
1      w  ■  m(m  -  1)      m{m  -  l)(m  —  2) 
12!  3! 

m(m  -  l)(m  -  2)Cm  -  3)  _ 
4  ! 
(1  -  w)(2  -  m)(3  -  m)  •••  (n  -  m) 
nl 


560  MISCELLANEOUS   EXAMPLES   VII. 

54.  Find  the  n*^  term  and  the  sum  of  the  first  n  terms  of  the 
series  i  +  3  +  7  4.  13  +  21  +  31  +  43  + -.. 

55.  Find  the  sum  of  the  first  n  terms  of  the  series 

3  +  4  a:  +  6  a;2  +  10  x3  +  18  x*  +  .... 

56.  Convert  a  +V2  +  a^  into  a  continued  fraction. 

57.  Convert  the  positive  root  of  ax^  —  ahx  —  6  =  0  into  a  con- 
tinued fraction. 

Denoting  by  Pi/Qi,  P2/Q2,  •••,  Pn/Qn,  the  successive  con- 
vergents  of  ai  +  hi 

a2  +  b2 


«3  +  ••• 
(Pi  =  «!,  ^1  =  1),  prove  the  following  relations : 

58.  P„  =  GnPn-l  +  bn-lPn-2, 
Qn  =  ClnQn-l  +  hn-lQn-2' 

59.  Pn+lQn  -  PnQn+l=(-  1)*'&1&2&3  -  K. 

60.  -Pw+l        ^»—(        2\n^l^2"'ftn 
Qn+l        Qn  QnQn+1 


61.   f^  =  ai  +  ^  -  -^^  +  -"  +(-  1)-  ^'^^ "'  ^"' 


Pn 


DETERMINANTS. 


661 


CHAPTER  XXXIX. 


Determinants. 


h 

c 

i 

c 

a 

1 

a 

h 

b' 

c' 

c' 

a' 

a' 

b' 

386.   If  the  matrices 

[Art  173.] 

be  used  to  denote  the  cross-product  differences 

6c'  —  6'c,  ca'  —  c'a,  ab'  —  a'b, 
the  values  of  x  and  y  derived  from  the  simultaneous 
equations 

ax  -^  by  +  c  =  0, 
a'x  +  b'y  +  c'  =  0, 
may  be  presented  in  the  form 


[Art.  172.] 


These  cross-product  differences  whose  matrices  have 
two  rows  and  two  columns,  are  called  determinants  of  the 
second  order. 

Ex.  1.  The  condition  that  the  equations 

ax  +  by  =  0,   a'x  +  b'y  =  0, 
may  be  simultaneous  in  x  and  y  is 


b     c 

c    a 

b'  c' 

,  y  = 

c'  a' 

a    b 

a    b 

a'  V 

a'  b' 

a      b 
a'   b' 


=  0. 


562 


DETERMINANTS. 


For,  if  the  equations  be  simultaneous,  then 


b' 
a' 


whence  (a6'  —  a'b)y  =  0, 

and  unless  ?/  =  0,  we  must  have  ab'  —  a'b  =  0. 

Ex.  2.   Write  down  the  solution  by  determinants  of  the  simul- 
taneous equations 

(C+Vy^  =  0,    3a; -y +  2  =  0. 
Solution  :     '^:. 


y  = 


2 

-1 

1 

2 

-1 

2  "  3 

-.1 

-1 

1 

1 

2 

2 

3 

• 

3 

-1 

4-1 


3-2^5 
1-6     7* 


Ex.  3.   Let  the  three  equations 

aix  +  biy  +  ciz  =  0, 

a2X  +  b2y  +  C2Z  —  0, 

azx  +  bzy  +  czz  =  0, 

be  regarded  as  simultaneous  in  x,  y.  z;  multiply  them  by 

biCz  —  &3C2,    &3C1  —  &1C3,    61C2  —  62C1, 

respectively  ;  add  the  resulting  equations  together ;  show  that  the 
new  coeflBcients  of  y  and  z  vanish  ;  and  thus  derive,  as  the  condi- 
tion of  simultaneity, 


ai 


62  C2 

-  a2 

bi   ci 

+  053 

6i  ci 

bz  cz 

63  Cz 

62  C2 

=  0. 


387.   When  its  three  matrices  are  replaced  by  their 
equivalent  cross-product  differences, 

62C3  —  63C2,  61C3  —  63C1,  61C2  —  52^1, 


DETERMINANTS. 


563 


expression 

tti 

62    c, 

—  0.2 

h    c. 

+  ^3 

61     Ci 

h         Cg 

h    C3 

^':i      C2 

becomes 

ttj^aCs  —  Cti&gCg  —  a2^iC3  +  ^26301  4-  tta^iCa  —  a362Ci- 
It  is  called  a  determinant  of  the  third  order,  and  its 
matrix  form  is 

ttj      61      Cj 
0,2     62     ^2 

ttg         63        Cg 

We  shall  frequently  denote  this  determinant  by  the 
abbreviated  notation  [^aib.jC^']. 

Let  it  now  be  required  to  solve  the  simultaneous  equar 
tions 

a^x  -f  biy  +  CiZ  +  di  =  0, 
a.^-\-b^  +  c,js  +  d2  =  0, 
«3^  +  ^32/  +  C32;  +  f^3  =  0. 

When  the  left  members  of  these  equations  are  multi- 
plied by 

62C3  —  63C2,  —  61C3  4-  &3C1,  and  biC^  —  b^i 

respectively,  and  the  resultant  equations  are  added 
together,  as  in  Ex.  3,  Art.  386,  the  new  coefficients  of 
y  and  z  become 

^1(^2^3  —  te)  —  h{biCs  —  fcgCi)  +  ^3(^1^2  —  &2C1)  =  0 
and 

^1(6263  —  ^3^2)  —  C2(&iC3  —  63C1)  +  c^(biC2  —  62C1)  =  0 

respectively,  and  there  remains 

0, 


«1 

&i 

c 

+ 

a2 

h 

C2 

tts 

h 

C3 

d, 

h 

Cl 

d. 

b,_ 

C2 

d. 

b> 

Cs 

664 


DETERMINANTS. 


whence,  in  the  abbreviated  notation, 

Eepeating  this  process  for  the  purpose  of  eliminating 
in  succession  z,  x  and  x,  y,  we  obtain 

[ai&sCg]' 

In  these  fractional  values  of  the  unknown  quantities 
we  observe  that  the  common  denominator  is  the  deter- 
minant [ai^a^s]?  and  that  the  numerators  are  three  deter- 
minants which  may  be  derived  from  [^162^3]  by  writing 
(in  the  matrix  form)  d^did^  in  the  place  of  aia2a^,  &i&2^3? 
and  C1C2C3  respectively. 

388.  A  determinant  is  fully  described  as  a  quantity 
when  its  terms  with  their  proper  signs  are  written  down; 
but,  for  the  purpose  of  discovering  its  properties,  it  is 
regarded  as  evolved  from  a  principal  term  of  the  form 
dib^Cgd^  •  •  •  Z„  by  keeping  the  letters  in  fixed  order  and  per- 
muting the  suffixes  1,  2,  3,  •  •  •  n  in  all  possible  ways,  or 
by  keeping  the  suffixes  in  fixed  order  and  permuting  the 
letters  in  all  possible  ways,  and  then  giving  to  the  terms 
thus  obtained  the  positive  or  negative  sign  according  as 
the  arrangement  of  the  suffixes  (or  letters)  in  such  term 
is  derived  from  that  of  the  principal  term  by  an  even  or 
an  odd  number  of  inversions  of  order ;  whereby  we  define 
that  an  inversion  of  order  occurs  as  often  as  a  suffix  (or 
letter)  precedes  another  suffix  (or  letter)  which  in  the 
principal  term  it  follows. 


DETERMINANTS. 


565 


For  the  determinant  of  the  third  order  this  method  of  evolving 
the  several  terms  is  indicated  in  the  following  scheme.  The 
letters  are  supposed  to  stand  in  the  order  a  be,  and  only  the  suffixes 
are  written  down. 


oral  order. 

1  inversion. 

2  inversions. 

3  inversions. 

Sign 

123 





__^ 

+ 

123 

132 





— 

123 

213 



. 

_ 

123 

213 

231 



+ 

123 

132 

312 



+ 

123 

132 

312 

321 

_ 

For  example,  there  are  three  inversions  in  3  2  1  because  3  pre- 
cedes both  2  and  1,  and  2  precedes  1. 


389.  Def.  A  determinant  of  the  n^  order  is  the  alge- 
braic sum  of  the  n  I  products  that  may  be  evolved  from 
a  principal  terra  aib^ajd^^'-lr^  by  keeping  the  letters  in 
fixed  order  and  permuting  the  suffixes  1,  2,  3,  •••n  in  all 
possible  ways ;  and  any  term  is  preceded  by  the  positive 
or  negative  sign,  according  as  it  is  derived  from  the 
principal  term  by  an  even  or  an  odd  number  of  inver- 
sions of  the  natural  order  of  the  suffixes  1,  2,  3,  •••  w. 

The  matrix  of  this  determinant  of  the  n^  order  is 


«1 

h 

cr-h 

ttg 

b. 

c,...k 

as 

63 

Cs-h 

• 

: 

;     i 

«„ 

K 

C.-ln 

This  determinant  is  frequently  denoted  by  the  abbrevi- 
ation [_aib2Cs " '  l^'] ',  sometimes  also  by  2  ±aib^s"'L' 

The  n^  quantities  aj,  bi,  Cj,  aj?  ^2i"'K  2,re  called  the 
elements  of  the  determinant,  the  n\  products  aA^^s •  •  •  ?„ , 


bm 


DETElRMINAN^S. 


cb^iC^"'ln,  etc.,  are  its  terms,  the  diagonal  line  aih^f^-'-l^, 
extending  downwards  from  the  upper  left-hand  corner, 
is  the  principal  diagonal,  and  the  product  ai&^Cs---^?  whose 
factors  are  the  elements  of  the  principal  diagonal,  is  the 
principal  term. 

In  the  present  discussion  we  shall  be  chiefly  concerned 
with  determinants  of  the  second  and  third  orders. 

390.  It  follows  from  our  definition  that  every  term 
of  a  determinant  of  the  n^^  order  contains  all  of  the  n 
letters  a,  6,  c,  ••/,  and  all  of  the  n  suffixes  1,  2,  3,  •••n, 
but  none  of  the  letters  or  suffixes  twice  repeated ;  hence, 
every  term  contains  one  and  only  one  element  from  each 
row,  and  one  and  only  one  element  from  each  column  of 
the  matrix. 


391.  If  in  a  matrix  of  a  determinant  the  letters  a,  b, 
c,"'l  and  the  numbers  1,  2,  3, ---n  be  interchanged, 
statements  concerning  rows  become  statements  concern- 
ing columns,  and  vice  versa;  hence,  a  determinant  in  the 
matrix  form  is  not  altered  by  changing  its  rows  into  col- 
umns and  its  columns  into  rows. 

Thus 


and 


Unless  the  contrary  is  explicitly  stated,  determinants 
will  henceforth  be  thought  of  as  given  in  the  matrix 
form. 


ai   &i 

= 

tti   a2 

0^2     &2 

bi    62 

J 

Oi  6i  Ci 

= 

tti  tta  as 

Ctg     ^2     ^2 

bi   62    ^3 

as 

h  c. 

Ci     C2 

C3 

'I 


DETERMINANTS. 


567 


392.  When  given  in  their  matrix  forms,  determinants 
of  the  second  and  third  orders  are  easily  expanded  into 
the  algebraic  sums  of  their  terms  by  the  rule  which  is 
expressed  in  the  following  formulae : 


a 

I  &i 

= 

=  aifta  ■ 

-  «2&1, 

» 

a^  62 

tti    61    Ci 

=  «! 

62  C2 

-a2 

&1    c, 

+  a3 

61    Ci 

«£     ^2     C2 

h    Cs 

&3     C3 

62  C2 

^3     &3     C3 

Ex.  1.  Expand 


2  3  1 
6  2   3 

3  12 


By  the  rule  of  expansion  this  determinant  is  equal  to 


2  3 
1  2 


3  1 
1  2 


+  3 


3  1 
2  3 


=  2(4 -3)- 5(6-1)+ 3(9 
=  2  -  26  +  21  =  -  2. 


2) 


Ex.  2.   Expand 


a 

h 

9 

h 

b 

f 

9 

f 

c 

=  A. 


By  the  rule  of  expansion  we  have 

A  =  aibc  -P)-h{hc  -  fg)  +  g{hf^  bg) 
=  abc  +  2fgh  -  aP  -  bg^  -  ch\ 


EXAMPLES    XCVIII. 

1.  Count  the  number  of  inversions  in  2143,  3421,  4321,  and 
53412. 

2.  If  abode  be  regarded  as  the  standard  order,  count  the  number 
of  inversions  in  acebd,  cdeba,  and  dbace. 


568 


DETERMINANTS. 


3.   What  are  the  signs  of  the  terms  hdk^  ahf,  and  Mc  in 


a 

b 

c 

d 

e 

f 

9 

h 

k 

4.  Write,  with  their  appropriate  signs,  all  the  terms  of  the  deter- 
minant [aih2Czd{\. 

5.  In  the  determinant  of  the  fifth  order  [^aih^Czd^e^l  determine 
the  signs  of  the  terms  asbiCzdse^,  aihsCidiCs,  a^h^C\d^e2^  and 
056403(^261  • 


Evaluate  the  following  determinants : 


6. 

1    2 
4   5 

3 
6 

7   8 

9 

• 

7. 

1 

-2 

3 

-2 

3 

-4 

3 

-4 

6 

8. 

1   3 
3   5 

5   7 

5 

7 
9 

9. 

1  0 
0  5 

3 
0 

7  0 

9 

. 

J 


10. 

— 

1 

0 

0 

0 

-1 

0 

0 

0 

-1 

11. 

0 

a 

b 

, 

h 

0 

a 

a 

b 

0 

• 

12. 

a 

b 

c 

b 

c 

a 

c 

a 

b 

. 

13. 

a 

b 

c 

b 

b 

b 

c 

b 

a 

. 

14.   Show  that 


as  bs  cs   =  — 

ai  bi  Ci 

= 

&i  ai  ci 

a2  ?>2  C2 

a2  &2  C2 

62  a2  C2 

ai  bi  ci 

as  bs  cs 

bs  as  Cs 

16.   Show  that 


1    -i 
i       1 


=  2,  where  i  =  y/—  1. 


DETERMINANTS.  569 

393.  Properties  of  Determinants.  An  interchange  of  any 
two  adjacent  rows  (or  columns)  cf  a  determinant  changes 
its  sign,  but  affects  its  value  in  no  other  way. 

For,  in  the  determinant  [aA^s •••?„],  the  interchange 
of  any  two  adjacent  rows  (or  columns)  merely  inter- 
changes two  adjacent  suffixes  (or  letters)  in  every  term, 
that  is,  produces  one  inversion  of  suffixes  (or  letters)  in 
every  term  without  disturbing  the  order  of  the  letters 
(or  suffixes),  that  is,  changes  the  sign  of  every  term,  and 
affects  the  terms  in  no  other  way. 

394.  An  interchange  of  any  tivo  rows  (or  columns)  of  a 
determinant  changes  its  sign,  but  affects  its  value  in  no 
other  way. 

For,  suppose  k  rows  to  lie  between  the  two  rows  that 
are  to  be  interchanged.  These  k  rows,  together  with  the 
two  rows  to  be  interchanged,  form  a  series  of  A:  +  2  rows. 
Then  we  may  make  the  interchange  by  passing  the  upper- 
most row  of  the  series  downwards  over  k  rows,  and  then 
passing  the  lowest  row  upwards  over  A;  + 1  rows.  The 
total  number  of  interchanges  of  adjacent  rows  is  there- 
fore 2k +  1,  an  odd  number,  and  each  such  interchange 
changes  the  sign  of  the  determinant.  Hence  the  total 
effect  is  a  change  of  the  sign  of  the  determinant. 

The  proof  for  the  effect  of  the  interchange  of  any  two 
columns  is  identical  with  the  foregoing. 

395.  A  determinant,  in  which  any  two  rows  (or  columns) 
are  identical,  is  equal  to  zero. 

For,  if  two  rows  (or  columns)  be  identical,  the  deter- 
minant is  unaltered,  either  in  sign  or  magnitude,  by  the 


570 


DETERMINANTS. 


interchange  of  these  two  rows  (or  columns).  But,  by- 
Art.  394,  the  interchange  of  any  two  rows  (or  columns) 
of  a  determinant  changes  its  sign.  Hence  the  determi- 
nant in  question  is  not  altered  in  value  by  changing  its 
sign,  and  its  value  must  therefore  be  zero. 

Ex.  1.   Determine  the  value  of 


1 

a    aP- 

1 

b    &2 

1 

c     c2 

Two  rows  of  this  determinant  become  identical  and  its  value 
zero  when  a  =  b.  Hence  it  must  have  a  —  b  as  a  factor  [Art.  148]. 
For  a  like  reason  it  contains  b  —  c  and  c  —  a  as  factors.  But  the 
determinant  is  of  the  third  degree  in  a,  6,  c.  Hence,  denoting  its 
value  by  A, 

A  =  Z(6-c)(c-a)(a-&) 

where  Z  is  a  number. 

But  the  principal  term  of  the  determinant  is  bc^,  and  this  is  the 
only  term  in  which  bc^  occurs,  and  the  coefficient  of  bc^  in 

L(b  —  c)  (c  —  a)  (a  —  b)  is  L  ;   .•.  L  =  1,    -, 
A=(&-c)(c-a)(a-6). 

Prove  that 

i37(l-/3)(^-7)(7-l)i 


ai37(tt-i3)(/3-7)(7-a). 


and 
Ex.  2. 


Ex.  3.    Prove  that 


1 

/3 

7 

1 

/32 

y2 

1 

^3 

7» 

/3     7 


./3« 


396.  If  all  the  elements  of  one  row,  or  of  one  column, 
of  a  determinant  be  multiplied  by  the  same  quantity,  the 
determinant  itself  will  be  multiplied  by  that  quantity. 


DETERMINANTS. 


571 


For,  every  term  of  the  determinant  contains  one 
element,  and  only  one,  from  each  row  and  from  each 
column  [Art.  390] ;  and  hence,  if  all  the  elements  of 
one  row,  or  of  one  column,  be  multiplied  by  the  same 
quantity,  every  term  and,  therefore,  the  sum  of  all  the 
terms,  will  be  multiplied  by  that  quantity. 

397.   A  determinant  in  which  any  tivo  rows,  or  any  two 

columns,  differ  only  by  a  constant  factor,  is  equal  to  zero. 

This  theorem  is  a  direct  consequence  of  Arts.  396,  397. 

For  example : 

0. 


ma  na  d 

=  mn 

a  a  d 

nib   nb   e 

b   b   e 

mc    nc  f 

c   c  f 

398.  Def.  When  any  number  of  columns  and  the  same 
number  of  rows  of  a  determinant  are  suppressed,  the 
determinant  formed  by  the  remaining  elements  is  called 
a  minor  determinailt. 

The  minor  determinant  obtained  by  suppressing  one 
column  and  one  row  is  said  to  be  of  the  first  order,  or  to 
be  a  first  minor;  similarly  the  result  of  suppressing  two 
columns  and  two  rows  is  called  a  second  minor ;  and  so  on. 

The  result  of  suppressing  the  column  and  row  through 
any  element  a;  of  a  determinant  A  is  called  the  minor 
corresponding  to  that  element  and  is  denoted  by  A,. 

Thus  the  first  minors  of  A  =  ai  b\  c\ 

Gi   62   C2 
as  ^3  •  cs 
corresponding  to  the  elements  ai,  61,  ci, 

respectively. 


&2     C2 

ai  C2 

aa   62 

Aa,= 

,    Aft  = 

,     Ac  = 

68    Cs 

\ 

as  Cs 

1 

as  6s 

572 


DETERMINANTS. 


It  is  evident  that  when,  in  the  first  row  or  first  column, 
all  the  elements  except  the  first  vanish  the  determinant 
reduces  to  the  product  of  the  non- vanishing  element  and 
its  corresponding  minor. 


ai  &i   Ci 

=  «! 

62     C2 

0       &2     C2 

63     C3 

0     63  Cs 

Thus 


399.  To  expand  a  determinant  of  the  n^^  order  in  terms 
of  n  of  its  first  minors. 

Since  every  term  of  the  determinant  [afiz'^s  •*•  ^n]  con- 
tains one  and  only  one  of  the  elements  a^,  ag?  ctg,  •••  a,^ 
[Art.  390],  we  may  write 

=  a^Ai  4-  ^2^2  +  0^3-^3  -\ 1-  a„ A?       («) 


a, 

61 

c,..-h 

012 

h 

Ci  —  k 

a. 

h 

Cs-ls 

a„ 

bn 

Cn-L 

ai 

bi  ci 

az 

62  C2 

as 

63  C3 

in  which  none  of  the  coefficients  Ai,  ^2^  •••  ^„  contain 
any  of  the  elements  ai,  a^^  •••  a„. 
For  example, 

ai(&2C3  —  63C2)+«2(&3Cl  —  61C3)  +  «3(&lC2  —  &2C1). 

The  quantities  A^,  Az,  -"  A^  are  called  the  co-factors  of 
ai,  ttg,  •••  a„  respectively.  We  require  to  determine  these 
co-factors  in  terms  of  the  minors  of  a^,  (Xg,  •••  a„. 

Por  this  purpose  let  the  severalrows  ag  \"'h^  %  ^3  ••*  ^sj 
•  •  •  a„  6„  •  •  •  ?„,  in  the  matrix  form  of  the  determinant 
[ai&2C3-"^„]?  be  alternately  moved  into  the  place  of  tho 


DETERMINANTS.  573 

first  row.  Then  the  determinant,  without  being  changed 
in  magnitude,  assumes  successively  the  forms 

...(_l)«-l[aAC2f^3-^n-l]; 

for,  when  any  row,  say  the  fc'**,  is  moved  into  the  place  of 
the  first  row,  it  passes  over  k  —  1  rows  above  it,  and  by 
such  a  transference  the  determinant  does  or  does  not 
change  sign  according  as  k  is  even  or  odd  [Art.  393]. 
Hence  (a)  may  have  the  following  several  forms : 

[aibiCsd^  •  •  •  Z„]  =  a^Ai  +  «2-42  H h  a„ A,       (a) 

-  [0261(53(^4  •  •  •  ^n]  =  dlA  +  ^2^2  +  —  +  a^A,  08) 

[036102^4  •••?„]=  ai^i  +  02^2  H h  a«  A,       (y) 

—  la^jbic^ds  •••?«]  =  tti^i  +  «2^2  H h  «« A>       (S) 

(_  1 )  -1  [a„6iC2  •  •  •  Z„-i]  =  cti^i  +  ^2^2  H h  «« A-       (^) 

Now,  in  (a)  put  02=  a3=  •••  =  a„  =  0 ;  then  [ai^a^s •  •  •/„] 
becomes  cii[62C3 •••?„],  that  is,  OiA^  [Art.  398],  and  there- 
fore 

ai^i=aiA^^. 

In  (/3)  put  ai=  a3=  04=  •••  =  a„=  0  ;  then  [a26iC3 •••?„] 
becomes  a2[6iC3  •••/„],  that  is,  a2^a,>  ^°^  therefore 

«2^2  =  —  <*2^a,- 

From  (y),  (8),  and  (X),  we  obtain  in  like  manner, 
03^3  =  as^^, 


674 


DETERMINANTS. 


Hence,    if    A    stand    for    the    original    determinant 

If  columns  and  rows  be  interchanged,  we  obtain 

A  =  aAa,  -  b,\  +  cAc, ( -  1)"-'^A^, 

or,  if  the  first  and  second  columns  be  interchanged, 

A  =  -  bA,^  +  ^2^5^  -  b,\  +...+(-  ^rK\ ; 

and  it  is  now  evident  that  the  expansion  may  be  made  in 
terms  of  the  minors  of  any  row  or  any  column. 

400.  If  the  elements  of  one  column  (or  row)  of  a  deter- 
minant be  multiplied  in  order  by  the  co-factors  of  the  corre- 
sponding elements  of  any  other  column  {or  row),  the  sum 
of  these  products  will  be  zero. 

In  the  determinant 


ai 

6i-..A;i... 

ag 

bz-'h"' 

«3 

bs"'ks"' 

let  the  elements  ki,  kz,  k^,'-'  be  multiplied  in  order  by 
the  co-factors  Ai,  A^,  A^,  •••of  the  elements  ai,  ag,  0^3,  •••, 
and  the  products  added  together.     The  result  is 

kiAi  -\-  k2A2  +  k^A^  +  •  •  •; 


DETERMINANTS, 
whose  matrix  form  is 


k. 

6,  •••*;,  ••• 

h 

hi  —  k^-- 

h 

h-h- 

575 


[Art.  399.] 


But  this  is  zero,  since  two  of  its  columns  are  identical. 
[Art.  395]. 

The  result  is  the  same  when  these  combinations  are 
made  with  any  two  columns  or  rows ;  for  any  column  or 
row  may  be  moved  into  the  place  of  the  first  column 
without  changing  the  magnitude  of  the  determinant. 

401.  If  each  element  of  any  column  (or  row)  be  a  bino- 
mial the  determinant  can  be  expi^essed  as  the  sum  of  two 
determinants  of  the  same  order. 

For,  if  Ai,  A2,  As,'" be  the  co-factors  of  the  elements 
of  the  first  column  of 

tti  -h  «!   bi  Ci  . 
0.2  +  "2    ^2    ^2    • 

ag  +  Og      63      Cg      . 


=  A, 


then,  by  Art.  399 

A  =  (aj  +  ai)^i  +  (a2  +  «2)A  +  (%  +  03)^3  +  — 

=  tti^i  -f  a2^2  +  «3^3  H h  «l^l  +  «2^2  +  (h^z  + 


tti  61  Ci 
0^2  ^2  ^2 
%     ^3     ^3 


+ 


«1  61  Ci  .  . 
"2  &2  Cg  .  . 
«3     &3     Cg      .    . 


576 


DETERMINANTS. 


402.  A  determinant  is  not  altered  in  value  by  adding  to 
the  respective  elements  of  one  column  {or  row)  the  same 
multiples  of  the  corresponding  elemeyits  of  any  other  column 
{or  roiv). 

For,  if  in  the  expanded  form 

aiAi  +  a2^2  +  a^As  +  ...  =  A, 

of  tlie  determinant 


a,  bi- 

..fc,... 

^2    h 

..fc^.-. 

a,   63. 

..^3- 

0     . 

• 

.     . 

. 

we  replace  ai,  as,  a^,  •••by  ai  +  mki,  ag  +  m7c2,  a^  +  mk^,"- 
m  being  any  number,  this  expanded  form  becomes 

(%  +  mki)Ai  4-  («2  4-  wfcg)  A  +  («3  +  'mks)As  H 

=  A  +  m  {kjAi  +  A;2^2  +  hA^  -\ ) . 

But  by  Art.  400 

kiAi  +  A;2^2  +  Ms  +  •••  =  0  ; 

ai-\-mki  bi"'ki" 
a2  +  mk.2  b2"'k2" 
a^  +  mk^  bs*"ks- 


Ex.  1.    Show  that 


1  1  3 
12  2 
1   5    -1 


=  0, 


DETERMINANTS. 


677 


By  adding  the  second  column  to  the  third  the  determinant 
becomes 

=  0. 


1   1   4 

=  4 

111 

1   2  4 

1   2   1 

1   5  4 

1   5   1 

Ex.  2.     Show  that       a       6   1 

h       a  1 

a  +  h  0  1 

By  adding  the  second  column  to  the  first  the  determinant 
becomes 

=  0. 


a  +  h  b  1 

=  (a  +  6) 

1    b  1 

b  -\-  a  a  1 

1  a  1 

a  +  b  0  1 

1   0  1 

Ex.  3.   Show  that     b  +  c  c  a  —  c    =  ab(b  —  c). 
2b     b  a-b 
b-^  a  a      a 

Subtract  the  second  column  from  the  first  and  add  it  to  the 
third  ;  the  determinant  becomes 


b   c    a 

=  ab 

1  c   1 

b  b    a 

1   b  1 

b  a  2a 

la  2 

Subtract  the  second  row  from  the  first ;  then 


ab 

1   c   1 
1   b  1 

1   a  2 

=:ab 

0  c-6  0 

1  b       1 
1      a      2 


=  ab{b  —  c). 


403.    To  express  the  product  of  two  determinants  {of 
the  second  order)  as  a  third  determinant  of  the  same  order. 

As  a  preliminary  step  in  the  demonstration,  we  tirst 
establish  the  following  lemma : 


578 


DETERMINANTS. 


If            ai  bi  m  n 

a2  h  p  q 

0  0  ai  A 

0  0  a2  ft 

then,  whatever  he  the  values  of  m,  n,  p,  q, 
A  =  Ai  •  Aa. 


=  A, 

a,    6i 

a2    h. 

ai    ft 

a2    ft 

=  A^ 


=  A, 


For,        'A  =  aj 


h  V  q 
0  ai  ft 
0     a2    ft 


a2 


6i  m  n 
0  ai  ft 
0      a2    ft 


=  a-J)2^2  —  (^^A2  =  ^1  •  ^2- 


Q.E.D. 


Now  in  A  put  m  =  q=  —1  and  n=p  =  0,  so  that 


ttj  6i  —  1 

a2  62     0 

0  0       ai 

0  0       a2 


0 
-1 

ft 
ft 


Then  multiply  the  first  two  rows  by  ai,  ft  respectively, 
and  add  the  products  to  the  third  row;  and  again, 
multiply  the  first  two  rows  by  a2,  ft  respectively,  and 
add  the  products  to  the  fourth  row.  The  result  of  these 
operations  is 


A  = 

a2 
«!«!  +  a2ft 
a^a-i  +  a2ft 

l>2 
&!«!  +  &2ft 
&1«2  +  &2ft 

-1      0 
0      -1 
0        0 
J       0        0 

= 
by  the  lemm 

-1      0 
0      -1 

a.     But 

aiCJa  +  a2ft 
-1       0 
0       -1 

&1«1  +  ^sft 
&1«2  +  &2^2 

=  1; 

DETERMINANTS. 


579 


cfri    6, 

«i    ft 

= 

aa    h 

«2       ft 

ax  h\  C\ 

and  A2  = 

ai   /3i   71 

Qi    62    C2 

a2   ^2   72 

as  &3  C3 

as   /33   73 

aicti  H-  cisft     &i<^i  +  ^2/?! 
aia2  +  ^2/^2     ^1*^2  4-  ^2ft 

Note. — This  method  of  proof  is  general,  and  may  be  applied  to 
determinants  of  any  order.  For  determinants  of  the  third  order 
the  result  is  as  follows  : 

If  Al: 


then 

A1A2  =    aitti  +  a2/3i  +  a37i  ?>iai  +  b^^i  +  b^yi  ciai  +  C2/3i  +  C371 

ai02  +  a2i32  +  0372    &ia2  +  &2i32  +  &372    Cia2  +  C2i32  +  C372 

aias  +  a2^3  +  0373   bias  +  b^^s  +  &373    Cias  +  c^^a  +  C373 

For  the  proof  in  detail  of  this  more  general  case  the  student  is 
referred  to  Smith's  Treatise  on  Algebra,  Art.  430. 

By  virtue  of  the  fact  that  a  determinant  is  not  altered  in  magni- 
tude by  changing  the  order  of  its  rows  or  columns,  or  by  inter- 
changing columns  and  rows,  the  product  of  two  determinants  may 
be  exhibited  in  various  forms. 


404.    To  solve  the  simultaneous  equations, 

aiX-\-biy-\-CiZ-\-dit  =  kif 
ttgX  +  b2y  +  cji  -\-  d^  =  kzf 
a-sX  +  63?/  +  c^z  +  d.t  =  ^3, 
a^x  +  642/  +  c^z  +  d^jt  =  ^4. 

Multiply  these  equations  in  order  by  A^,  A^,  A^,  A^,  the 
co-factors  of  cii,  a^,  a^,  a^  respectively  in  the  determinant 
laib./^sd^'],  and  add  the  resulting  equations  together.  Then, 
by  Art.  400,  the  new  coefficients  of  y,  z,  and  t  vanish,  and 
there  remains 

^a^Ai  -f  ttaA  +  «3^3  +  (iiAi)tc  =  kiAi  -h  A;2^2  +  hA^-^k^A^, 

that  is,  laib2Cs,d4]x  =  Ikib^c^d^]. 


580 


DETERMINANTS. 


By  using  as  multipliers  the  co-factors  of  b^,  62?  ^3?  ^4, 
we  obtain  in  a  similar  manner 

and  the  symmetry  of  these  relations  shows  that  we  shall 
have  for  the  values  of  z  and  t 

laib2Csdi\z  =  [aib^hd^'], 

[0162^3(^4]^  =  \^aib2C.Jc^']. 

This  method  is  obviously  general,  and  readily  appli- 
cable to  simultaneous  equations  with  any  number  of 
unknown  quantities. 

405.  Eesultants.  A  resultant,  or  an  eliminant  as  it  is 
also  called,  is  a  determinant  the  vanishing  of  which  is 
the  necessary  and  sufficient  condition  in  order  that  two 
or  more  equations  may  be  satisfied  by  the  same  value  or 
values  of  the  unknown  quantities  involved.  Thus  the 
resultant  of  the  equations 

ax-\-by  =  0, 

a'ic  +  6'?/  =  0 

is  aV  —  a'&  [Ex.  1,  Art.  386],  and  the  resultant  of  the 

equations 

a^x  +  b^y  -f  c^z  =  0, 

ttgic  +  boy  +  c^z  =  0, 

o^^  +  hy  +  C32;  =  0, 

is  tti  bi  Ci 

C(/2     O2     C2 
^3     Oq     C3 


[Ex.  3,  Art.  386.] 


DETERMINANTS. 


581 


Ex.  1.   Find  the  resultant  of  the  equations 
aic2  +  6x  +  c  =  0, 
a'«2  +  b'x  +  c'  =  0. 

Multiplying  each  equation  by  x  we  obtain  the  set  of  four  equa- 
tions. 

ax^  +  6x2  +  ex  +  0  =  0, 

0  x8  +  ax2  +  6x  +  c  =  0, 

a'x^  +  6'x2  +  c'x  +  0  =  0, 

0x»  +  a'x2  4-  b'x  +  c'  =  0, 

which  may  be  treated  as  simultaneous  equations  in  the  four  quan- 
tities x**,  x2,  X,  1. 

Multiply  these  equations  in  turn  by  the  co-factors  of  the  fourth 
column  of  the  determinant  formed  by  the  coefficients  a,  b,  c,  0,  etc., 
and  add  the  resulting  equations  together.  Tlien,  by  Art.  400,  the 
new  coefficients  of  x*,  x^,  and  x  all  vanish  and  there  remains 


a  6  c  0     =0. 

0  a  6  c 

a'  h'  c'  0 

0  a'  V  d 


This  determinant  is  the  resultant  required  and  its  vanishing  is 
the  condition  necessary  (it  may  also  be  proved  sufficient)  in  order 
that  the  equations 

ax2  -f  6x  +  c  =  0,    a'x2  +  6'x  -f  C  =  0 

may  have  a  common  root.     [Compare  Ex.  3,  Art.  173.] 

Ex.  2.   Find  the  condition  necessary  and  sufficient  in  order  that 

the  equations 

ax8  +  6a;2  +  ex  +  (i  =  0, 


px2  +  gx  +  r  =  0 


may  have  a  common  root. 


682 


DETERMINANTS. 


Multiplying  the  first  equation  by  x  and  the  second  by  cc^  and  x 
we  obtain  the  set  of  five  equations 

ax^  +  hx^  +  cx^  -{-  dx         =0, 

ax^  +  6x2  +  cic  +  (?  =  0, 

px^  -\-  qx^  +  rx^  =  0, 

px^  +  qx'^  -\-  rx        =0, 

•  px^  +  qx-\-r  =  0. 

If  these  are  to  be  satisfied  by  the  same  value  of  x  we  must  have 

a  h  c  d  (i    =0, 

Q  a  h  c  d 

p  q   r  0  0 

0  p  q  r  0 

0  0  p   q  r 

which  is  the  required  condition. 

The  method  of    elimination  employed  in  these  examples  is 
called  Sylvester's  dialytic  method. 


EXAMPLES  XCIX. 

1.  Transform  a  h  c 
d  e  f 
g    h  k 

into  an  equivalent  determinant  having :  (1)  de/as  its  first  column, 
(2)  cfh  as  its  first  row,  (3)  adg  as  its  third  row. 

2.  Solve  by  determinants 

3x  +  2?/  +  2  =  0, 
5x+12y-l=0. 

3.  Solve  by  determinants 

5x  +  y  -  40-3=0, 
2x4-3^  +  70  +  4  =  0, 
3x -22/+ 50- 7=0. 


DETERMINANTS. 


583 


4.  Find  a  value  of  k  such  that  the  equations 

2a;-3y  +  l  =  0, 
Zx  -  by  -\-k  =  (i, 
x  +  y  -7  =0, 
may  be  simultaneous  in  x  and  y. 

5.  Solve  by  determinants 

x-2y-3z  -\-2t  =  2, 
5x-{-y-\4tz  +  6t  =  3, 
6x-Sy  -h  z+  10«  =  0, 
7x-2y-\-2z  +  2t=-2. 


Evaluate  the  follov\ring  determinants 
6.     8  2  8  10. 

11 


8 

2 

8 

7 

5 

6 

9 

4 

1 

a  —  h 

X 
X 


X 

b-c 

X 


X 

X 

c  —  a 


0 

h 

9 

h 

0 

f 

9 

/ 

0 

11. 


1  4-a 
1 

1 


1 

1  +  6 

1 


1 

1 
1  +  c 


9. 


14. 


4     3     7 

2 

18. 

1  a  a2  a8 

6     0     2    1 

1   6  62   6« 

6-210 

1  c  c2    c8 

3     6     4  3 

' 

1    d    (22    (^ 

• 

X       a     —  b 

13. 

b  —  c     c          b 

—  ax        c 

-  c     c  —  a      a 

b      —  c      X 

• 

-  b    —a     a  —  b 

le  f oUov^ing  ide 

ntities : 

b-\-  c 

a  -  c 

a-b 

=  8  abc. 

b-c 

c-i-  a 

b-a 

c  - 

-b 

c  -  a 

a  -\-b 

584 


DETERMINANTS. 


15. 


16. 


17. 


18. 


&+c+2a         h  c 

a        c  +  a  +  26         c 
a  b        a  +  b  +  2c 


=  2(a  +  6  +  c)8 


aixi  +  biX2  aivx  +  bxy^ 
a^xi  +  b^Xi  a^yi  +  &22/2 


1    X   x^. 

• 

1  2/  2/2 

1     5?    2!2 

a\  a^  as 

6i     62     &3 
Ci     C2     C3 

and  therefore, 

ai  bi   ci  2 

^2  ^2  C2 
as  bs  cs 


Ai  Bi   Ci 

= 

A2  Bi    Ci 

As  Bz    Cs 

aixi  +  biyi  aiX2  +  &12/2  I 

a2a:i  +  &22/I    «2a;2  +  ^22/2  I . 

(w  +  x)2  (v  +  xy  (w  +  a-) '2 

(W  +  0)2    (V  +  0)2    (to  +  0)2 


«!  ^2  «3 
61  62  &3 
Ci     C2     C3 


^1  ^1    Oi 

A2  B2  C2 

^3    -B3     O3 


where  ^1,  J5i,  ^2?  etc.,  are  the  co-factors  of  ai,  5i,  a2,  etc.,  in  the 
determinant  [ai&2C3]. 

19.  If  X  +  y  +  0  =  aic  +  6y  +  C0  =  0,  prove  tl^^t 


X  y  z    =0. 
cab 
b  c  a 


20.  Find  the  condition  necessary  and  sufficient  in  order  that 

ax2  +  6x  +  c  =  0  and  x^  =  l 
may  have  a  common  root. 

21.  Determine  X  and  Fin  terras  of  «,  6,  x,  and  y,  such  that 


a  6 

• 

a;  2/ 

= 

X  Y 

b  a 

y  X 

Y  X 

RATIONAL  FACTORS  AND    HIGHER   EQUATIONS.    585 


CHAPTER  XL. 

Rational  Factors  and  Higher  Equations. 

406.  To  Factor  7? -\-f  +  :^ -Zxyz.  In  Art.  126  it  was 
seen  that  this  ternary  cubic  is  identically  equal  to 

(«  +  ?/  +  z)  {x^  +  if  +  z^-yz-zx-  xy\ 

so  that  one  of  its  linear  factors  is  x  -\-y  -\-z.  Arrange 
the  terms  of  the  second  factor  as  a  quadratic  in  x,  and 
add  and  subtract  {y  -\-  z)y4,  thus : 

a;2_  (,jj^z)x  +  ^1±^  -  {y-±ll  +  f+z'-yz.      [Art.  122] 
In  more  condensed  form  this  is 

or,  as  the  difference  of  two  squares, 

(-^)'-e^-T)" 

Its  two  factors,  rational  and  linear,  in  x,  y,  z,  are  now 
easily  written  down  [Art.  115].  When  simplified,  by 
collecting  coefficients,  they  are 

x-\-  (x)y  -\-  <D%  x  -\-  (Jy  -\-  (oZf 


586  EATIONAL  FACTORS   AND  HIGHER  EQUATIONS. 

where,  for  brevity,  w  and  a>^  are  written  in  place  of 
-i  +  iV^=^  and  -^-^V^ 

which  the  student  will  recognize  as  the  two   complex 
cube  roots  of  unity.     [Art.  188.] 

Hence  the  three  rational  factors  of  the  first  degree  of 
Q^-^-f-^-z'^—Zxyz  are  x  +  y-{-z,  x  +  o}y-\-(o%  x+oy^y+wz, 
where  to  and  o>^  are  the  complex  cube  roots  of  unity. 

407.   To  Pactor  the  Cubic  in  x:    To  factor  the  special 
cubic   in  a;,  of  the  form  a^+qx-\-r,   add   and   subtract 
—  y^  —  z^  —  SxyZf  producing  the  identity 
a^ -{-  qx  -\-  r  =  a^  —  y^  —  :^  —  S xyz  +  y^+  :^ -\-xCSyz-hq)  +  r 
and  determine  y  and  z  by  the  conditions 

.    f  +  ^  =  —  r,Syz  =  —  q.—' 

Then  at  once  [Art.  406] 

a^  +  qx  +  r  =  a:^  —  y^  —  ^  —  S  xyz 

=  (x  —  y  —  z)(x  —  wy  —  mh)  (x  —  <i?y  —  ws;), 

where  w,  <o^  are  the  complex  cube  roots  of  unity. 

From  the  two  equations  in  y  and  z  is  obtained,  by 
eliminating  z^  the  quadratic  equation 

or  2/6^^^_|_^0^ 

whence  2^  =  _r  +  ^(J  +  g, 


RATIONAL   FACTORS   AND   HIGHER   EQUATIONS.    587 

408.  The  general  cubic  expression  x^  -\-  p7?  -\-  qx-\-r 
may  be  reduced  to  the  special  form  y^  -\-  Qy  -\-  B.  By 
easy  reductions  we  obtain  the  identity 

(y  +  hy  ^p(y-^hy-^q(y  +  h)-\-r  =  f-{-(p  -[-Sh)y' 

^(q-\-2ph-^Sh^y^r-\-qh-^ph^-\-  h\ 

The  coefficient  of  y^  will  vanish  if  h  =  -^^p'^  hence 
the  substitution  x  =  y  —  \p  will  effect  the  required  trans- 
formation.    The  new  coefficients  are 

R  =  r-\pq-\-^p\ 

and  the  identity  sought  for  is 

y?^px'J^qx  +  r={x  +  \pf+Q{x-\-\p)  +  R. 

The  factors  of  this  general  cubic  may  now  be  written 
down  in  accordance  with  the  formula  of  the  last  article. 
Thus 

a?-\-p7?+qx-\-r 

=  (x-\-\p-T-Z)(x-\-\p-iaY-u?Z)(x-\-\p-iJ^T-ioZ), 


where  r«  =  -:?  +  A/^'  +  ^ 


2"^VT'^27' 


Z3  =  -^-J^V^. 
2      A' 4  ^27 

409.   The  three  factors  of  the  last  article  yield  at  once 
the  roots  of  the  cubic  equation 

Q?  +  p7?  -f  ^-a;  +  r  =  0. 


588   RATIONAL  FACTORS   AND   HIGHER   EQUATIONS. 
These  roots  are 

x  =  -lp-\-<oY-j-(x)^Z, 

When  p  —  0,  the  first  of  these  expressions  for  the 
value  of  X  becomes 

which  is  known  as  Oardan's  Pormula,  for  the  solution  of 
the  equation 

410.   To  Factor  the  Biquadratic  in  x :    Let  it  be  assumed 
that  the  general  biquadratic 

s^-^pof  +  qx^  -^rx±s 

can  be  expressed  as  the  difference  of  two  squares  of  the 
form 

(x^  +  ax-j-ipy-(bxA-oy, 

and  let  the  values  of  a,  6,  c,  p  be  sought  which  will  make 
such  a  reduction  possible.  Expanded  and  arranged  ac- 
cording to  the  powers  of  x,  this  difference  becomes 

a;*  +  2aaj3  +  (a^  -f  p  -  62)ic2  4.  (^^  _  2  bc)x-\-  ip'  -  c^. 

This  will  be  identical  with  the  given  biquadratic  expres- 
sion if 

2a=p,     a^  -\-  p  —  b"^  =  q, 

ap  —  2bc  =  ry         ^p^  —  (^=:s, 


BATIONAL  FACTORS   AND   HIGHER  EQUATIONS.    589 

and  these  four  equations  are  sufficient  to  determine  a,  h, 
c,  and  p.     Eliminating  a,  we  have 

2bc  =  ^pp  —  r,  4:b^=p^  —  4:q  +  4:p,  c^  =  \p^  —  s] 

whence   {p^  —  4:q  -\-  4: p)  (\ p^  —  s)  —  (^pp  —  rf  =  0, 

or         p^  —  qp^ -{- (pr  —  4:S)p—p^s-\-4:qs —  r^  =  0. 

Thus  the  solution  of  a  cubic  equation  (the  reducing  cubic 
it  is  called)  is  a  necessary  part  of  this  transformation. 
When  one  root  of  this  cubic  equation  is  found,  the  values 
of  a,  b,  and  c  are  at  once  known,  and  the  biquadratic 
form  is  then  expressible  as  the  product  of  the  two  quad- 
ratic factors 

lay^^(a-\-b)x  +  ip  +  cl'\x'-\-(a-b)x-{-ip-c\,   . 

each  of  which  may  be  resolved  into  the  product  of  two 
factors  of  the  first  degree  by  the  method  of  Art.  122. 

411.   The  roots  of  the  biquadratic  equation 

iC*  -i-pa^  +  qa^  +  ^^  +  ^  =  0 
are  now  obtained  by  solving  the  two  equations 

a^  4-  (a  4-^)2;  +  ^p  +  c  =  0, 

a^-{-(a-b)x  +  ^p-c  =  0. 
These  roots  then  are 


x  =  -^(a  +  b±V(a^y-2p-4:c)y 

oc  =  -^(a-b±  V(a-by-2p-\-4:c), 

a,  6,  c,  and  p  being  obtained  by  means  of  the  equations 

P^  —  QP^  +  (P^  —  4:  s)p—ph-\- 4.  qs  — 7^  =  0, 

a  =  ^p,   6  =  ^Vi>^  — 4g  +  4p,   c  =  iV/3^  — 4s. 


690    EATIONAL  FACTORS   AND    HIGHER   EQUATIONS. 

The  general  algebraic  solution,  in  terms  of  radicals,  of 
equations  of  higher  degree  than  the  fourth  has  been 
shown  to  be  impossible. 

.EXAMPLES  C. 

Solve  the  following  cubic  equations  : 
,^  1.   2  x3  +  X  -  3  =  0.  ^^  4.   a;3  +  12  5c  -  12  =  0. 

2.   x^-60x-  189  =  0.  6.   x^-6  a^x  -  9  a^  =  o. 

^3.   x^-Qx-Q  =  0.  6.   x3-6ax2-6a%-2a3  =  0. 

Transform  the  following  expressions  into  the  products  of  quad- 
ratic factors :  > 

7.  9x4-27x3  +  18x-4. 

8.  X*  +  4  x3  +  4  x2  -  5  X  -  12. 

9.  x*  -  3  x3y  +  2  xV  -  »^?/^  -  2/*» 

10.  x^  +  4  x^  -  2  x2  -  12  X  -  6. 

11.  4  x*  +  8  x^i/  -  4  xr/3  -  yK 
12.   Show  that 

ax3  +  3  hx^y  +  3  cxy'^  +  dy^  =p(x  +  XyY  +  g(x  +  fiyy, 
..  .  1     F  +  ^A      1     -F-yA 


and     p,  g  =  K«  +  ^«^-4J2'VA)»    i(a-Va2_4^V^); 

where  i^,  H=ad  —  be,    ac  —  &2, 

and  A  =  (ad  -  5c)2  -  4  (ac  -  62)  (bd  -c^). 

412.   To  Factor  the  General  Quadratic  inx,  y: 
aa:^-\-2hxy  +  %'  +  2^a;+  2/^  +  c. 

This  expression,  arranged  according  to  descending  powers 
of  X  and  multiplied  by  a,  has  the  form 

(axy-^2{hy-\-g)ax  +  a(by'  +  2fy-{'C). 


BATIONAL  FACTORS   AND   HIGHER   EQUATIONS.    591 

It  may  here  be  regarded  as  a  quadratic  in  arc,  and  may  be 

factored  by  the  method  of  Art.  122.     Thus,  adding  and 

subtracting  (Tiy  -\-  gfy  ^^  ^^^^ 

{axf  ■\-2Qiy^  g)  ax  +  Qiy  +  gf-  (hy+gy+a(by'-h2fy-\-c) 
=  (ax  +  hy  +  gy  -  IQi'  -  ah)  f -2 (af-  hg)y  +  ^^ _  ^^ 
=  {ax  ■i-hy-\-g  +  ^D)  (ax-{-hy  +  g  —  y/D), 

where  D  stands  for  (h^  —  ab)  y^  —  2  (af—  hg)  y  -\-g^  —  ac. 

If  these  factors  are  to  be  rational  integral  functions  of 
X  and  y  of  the  first  degree,  D  must  be  a  perfect  square, 
the  condition  for  which  is 

(h^  -  ab)  if  -  ac)  -  (af  -hgy  =  0,     [Art.  179] 

or  a^hc  +  2  afgh  —  a^f^  —  abg^  —  ach^  =  0 ; 

or,  since  a  is  not  zero, 

abc  +  2fgh-af  -  bg'-ch^  =  0. 

It  is  easy  to  verify  that 

abc  +  2fgh  -  af  -  bg^-ch^  = 

[See  Art.  385,  Ex.  2.] 

This  expression  (function  of  the  coefficients)  is  called 
the  discriminant  of  ax^  -\-  2  hxy  +  •  •  •  +  c,  and  is  usually 
denoted  by  A.  The  first  part  of  our  conclusion  now  is : 
If  aa^  4-  2  hxy  +  •  •  •  +  c  is  resolvable  into  linear  factors, 
then  necessarily  A  =  0. 

But  conversely,  if  A  =  0,  then,  taking  positive  values 
only  of  V/i^  —  ab  and  ^/g^  —  ac, 


a 

h 

9 

k 

b 

f 

9 

f 

c 

af—hg  =  V/i^  —  ab  •  -Vg^  —  ac,  if  af—  hg  >  0, 
(af—  hg)  =  -y/h^  —  ab  •  VgT—aCf  if  af—hg<,  0. 


592  EATIONAL  FACTORS   AND   HIGHER   EQUATIONS. 
In  the  former  case 


in  the  latter 

D  =  ( V^'  -  ah  .  y  +^g^-aG)\ 

and  a  (ax^  +  2  hxy  -\-  by^  -\-  2  gx  +  2fy  +  c) 


is  (ax  +  %  4-  gy  —  (Vh^  —  ab  •  y  —  -Vg^  —  ac)^, 
if  af—  hg  >  0, 

but  is  (ax  +  hy  -\-  gY  —  ( V/^^  —  ab  •  y  +  V^^  —  ac)^, 
if  af—hg<0. 

These  differences  of  squares  are  immediately  resolva- 
ble into  the  product  of  rational  factors  of  the  first  degree 
in  X  and  y.     Our  result  is : 

A  rational  integral  function  of  the  second  degree  in  x,  y 
may  be  resolved  into  the  product  of  two  rational  integral 
factors  of  the  first  degree  if  and  only  if  the  discriminant 
be  zero. 

Carried  to  its  conclusion,  the  factoring  process  gives 
us  the  formula 

ajx'  +  2hxy  +  by''  +  2gx  +  2fy  +  G 

=  -  [ax  +  (^  +  V/i^  —  ab)y-\-gT  '\/g^  —  ac] 


X  [ax 4-  (h—-\/h^  —  ab)y  +  g  ±  V/  —  ac], 

in  which  the  upper  signs  are  to  be  taken  if  af—hg>  0, 
the  lower  if  af—  hg  <  0. 

Ex.  1.   Factor  2x^  -  xij  -  y^ -{- Sy -2. 

Here  h^  —  ab  =  f ,  g"^  —  ac  =  4,  af  —  hg  =  3.     The  upper  signs 
are  to  be  taken,  and  the  factors  are  2  aj  +  y  —  2,  x  —  ?/  +  1. 


RATIONAL  FACTORS   AND   HIGHER   EQUATIONS.    593 

Ex.  2,   Factor  2  x'-^  -  2  icy  -  y^  -  2  x  -  2  y  -  1. 
Here  Ti^  -  ab  =  S,  g"^  -  ac  =  S,  af— hg  =  —  3.     The  lower  signs 
are  to  be  taken,  and  the  factors  are 

(V3  +  l)aj  +  2/  +  l,  iy/S-l)x-y-l. 

EXAMPLES  CL 

Find  the  factors  of  such  of  the  following  expressions  as  are 
factorable  ; 

1.  x2-2/2_aj  +  3y-2. 

2.  2x^-2y^-x-Sy-l. 

8.  2  x2  -  3  xy  -  2  1/2  4-  3  x  +  4  y  -  2. 

4.  x^  +  xy-y^  +  Sx-1. 

6.  x'^-4xy  +  Sy^  +  2x-2y. 

6.  2x^-7xy  +  Sy^-3x~y-2. 

7.  2x^  +  bxy-}-2y^-x  +  y-l. 

8.  2x^-Sxy-2y^  +  x-\-lSy-16. 

9.  x2  -  y2  +  (1  _  c)  x  +  (1  +  c)  y  -  c. 

10.  2  x2  +  2  xy  +  y2  +  6  a;  +  6  y  +  9. 

11.  x2  -  xy  +  2/2  -  X  -  y  4-  1. 

12.  2x2-4xy +  y2_2/y-/2. 

Determine  the  value,  or  values,  of  c,f,  a,  and  g  which  will  make 
the  following  expressions  separable  into  factors  of  the  first  degree 
in  X  and  y  : 

13.  x2-y2_7a;  +  9y-c. 

14.  x2  -  5  y2  +  3  a;  +  2/y  +  1. 

15.  grx2-y2+25rx-3y-l. 

16.  0x2  -  2  xy  +  3  y2  _  X  4-  2  y  +  2. 

17.  ax2  +  2  xy  -  y2  +  3  X  -  4  y  -  a. 

18.  Find  the  relation  between  /  and  b  which  will  make 
x2-by^  +  Sx-}-2fy  +  l 
resolvable  into  factors  of  the  first  degree  in  x  and  y. 
2q 


594   RATIONAL  FACTORS   AND   HIGHER   EQUATIONS. 

413.  Simnltaneoiis  Equations.  If  each  of  a  set  of  simul- 
taneous equations  of  higher  degree  can  be  factored,  their 
solution  may  be  reduced  to  that  of  a  set  of  simultaneous 
equations  of  lower  degree.     Suppose  that 

are  simultaneous  equations  in  x  and  y,  and  that  <^  and  \f/ 
can  be  resolved  into  factors  of  the  first  degree,  say 

<l>  =  l'm'n,    i{/  =  p'q. 

Since  </>  and  j/r  are  simultaneously  zero  if,  and  only  if,  a 
factor  in  each  is  zero  [Art.  127],  all  the  solutions  of 
cfi  =  0,  \l/  =  0  will  be  obtained  by  solving,  in  turn,  the  fol- 
lowing pairs  of  equations  of  the  first  degree : 

Z  =  0,  j9  =  0;     m  =  Oj  p  =  0'j     n  =  0,  p  =  0; 
1  =  0,    q  =  0'j     m  =  0,    q  =  0'j     n  =  0,    q  =  0; 

and  obviously  there  can  be  no  others.  The  two  groups 
of  equations,  the  <^-j/^  group  and  the  l-m-n-p-q  group,  are 
called  equivalent  systems. 

Def.  Two  systems  of  equations  are  equivalent  when  all 
the  solutions  of  either  are  identically  the  same  as  those  of 
the  other. 

Ex.  1.  Solve  the  simultaneous  equations  ic^  _  ^2  _  2  a;  -j- 1  =  0, 
x2  -  4a;^  +  3?/2  +  X  -  31/ =  0. 

The  factors  of  the  first  are  x  +  y  —  \  and  x  —  y  —  1]  of  the 
second,  x  —  ^y  and  cc  —  y  +  1.    Hence  an  equivalent  system  is 
(1)  a;  +  1/  -  1  =  0,        x-^y  =  0 
(2)x  +  y-l=0,    x-y-\-\  =  (i 

(3)  x-y  -1=0,        x-Sy  =  0 

(4)  X  -  y  -  1  =  0,    x-y+l  =  0 
the  solutions  of  which  are  respectively 

a;,  y  =  i  i  ;  0,  1 ;  I,  i ;  and  00,  co. 


RATIONAL  FACTORS   AND   HIGHEB   EQUATIONS.    595 

Ex.  2.   Solve  the  simultaneous  equations 

z^  +  y^  -  10  =  0,   x^-2xy  -{-y^-2x  +  2y  =  0. 

Since  x^  -  2 xy  +  y"^  -  2x  +  2 y  =  (x  -  y  -  2) (x  -  y),  an  equiva- 
lent system  here  is 

(1)  x-y-2  =  0,    x2  +  2/2_io  =  0;       * 

(2)  x-y  =  0,    x2  +  y2_io  =  o. 

The  solutions  are  easily  found  to  be : 

for  the  first  set,       x,  y  =  3,  1,  or  —  1,  3 ; 

for  the  second  set,  x,  y  =  y/6,  y/6,  or  —  ^^5,  —  ^5. 

414.  The  foregoing  method  of  solving  a  pair  of  simul- 
taneous equations,  <^  =  0,  «/'  =  0,  fails  unless  either  <f>  or 
i}/,  or  both,  can  be  resolved  into  integral  factors  of  the 
first  degree,  and  it  is  not  usually  true  that  an  integral 
function  of  the  second  (or  higher)  degree  in  x  and  y  has 
such  factors. 

When  <^  and  ij/  are  both  of  the  second  degree,  however, 
it  is  always  possible  to  find  a  quantity  k,  such  that  the 
quadratic  function  (fi-\-k\p  shall  be  resolvable  into  two 
factors  of  the  first  degree.  For  this  purpose  we  have 
only  to  put  the  discriminant  of  <^  +  kx^/  equal  to  zero  and 
solve  the  resulting  equation  in  k  [Art.  412].  We  then 
substitute  the  system  <^  =  0,  <^  +  fci/^  =  0,  for  the  system 
<^  =  0,  i/^  =  0. 

Suppose  a  value  of  k  found  for  which  the  discrimi- 
nant vanishes ;  then 

<t> -\- kip  =  I  •  m, 

where  /,  m  are  of  the  first  degree,  and  the  system  of  two 
sets, 

(1)  <^  =  0,  Z  =  0;     (2)  <^  =  0,  m  =  0; 

is  now  equivalent  to  the  one  set,  <^  =  0,  ^  =  0. 


596   RATIONAL   FACTORS   AND   HIGHER   EQUATIONS. 

Let  us  apply  the  method  to  an  example. 
Ex.  1.    Solve  x"^  +  2xy  -  y"^  -  2 X  +  2 y  +  1  =  0, 

a;2  +  2/2  _  5  =  0. 

The  function  <(>  +  k\}/  in  this  case  is 

x^  +  2xy  -  y^  ~  2x  +  2 y  +  1  +  k(x^  +  y'i  -  5), 
or        (i^k)x'^  +  2xy  +  (k-l)y'^-2x  +  2y  +  (il-5k); 
and  its  discriminant  is 

(A;  +  1)(A;  -  1)(1  -  5A:)- 2 -(1  +  A;)-(A:- !)-(!  -  5A:), 
or  -^k^^k^-^-Sk-i. 

This  is  zero  if  A;  =  1.     For  this  value  of  A:,  0  +  k<p  becomes 
2x'^  +  2xy  -2x  +  2y  -4, 

or  in  its  factored  form, 

2(x  +  l)(x  +  y-2). 

Our  problem  is  now  reduced  to  the  solution  of 
x  +  1  =  0,    x2  +  y2  _  5  =  0, 
from  which  we  find 

x,y=-l,2,  or  a;,  y  =  -  1,  -  2, 
and  of  X  +  y  -  2  =  0,   a;2  +  2/2  _  5  _  0, 

whose  solutions  are 

x,y  =  l  +  ^V6,  1-^V6,  or  x,y  =  l-^y/6,  l  +  jVC. 

Ex.  2.   Solve  ic2  +  2/2  +  4  a;  -  1  =  0, 

a;2  _  ^2  _  2  y  _  1  =  0. 
The  factored  form  of  ic2  —  2/2  —  2  2/  —  1  is 
(x-y-l)(x  +  y  +  l), 
and  the  discriminant  of 

ic2  4-  2/2  +  4 X  -  1  +  A:  (x2  -  2/2  -  22/  -  1) 
is  O.F  +  O.A;2  +  3A;-6. 


RATIONAL  FACTORS   AND  HIGHER  EQUATIONS.    597 

Hence  (x  —  y  —  l)(x  +  y  -\- 1)  =  0, 

r.2  ^  y2  ^  4x  -  1  +  ^(x^  -  y^  -2y  -  1)  =  0 
is  an  equivalent  system.    The  last  equation  reduced  is 

4 a;2  _  2/2  +  6x  -  5 2/ -  4  =  0, 
or  in  its  factored  form, 

(2x  -  y  -  l)(2x  +  y  +  i)  =  0. 
Hence  the  following  is  another  equivalent  system : 

(l)ic-y-l  =  0,  2a;-2/-l=0;  x,y  =  0,-l; 

(2)  x-y-l  =  0,  2x-\-y-\-i  =  0;  x,y  =  -l,-2; 

iS)  x  +  y+\l  =  0,  2x-y-l=0;  x,y  =  0,  -1; 

(4:)x  +  y  +  l=0,  2x-\-y  +  4  =  0;  x,y  =  -3,2. 

The  solutions  are  written  at  the  right  of  each  pair  of  equations. 

EXAMPLES  OIL 
Solve  the  following  systems  of  simultaneous  equations : 

1.  jc2+2«2/  +  y2-3x-3y+2=0,  2x:2-Sxy-\-y^-5x-\-3y-\-2=0. 

2.  Sx^-{-Sxy-Sy^=0,  ix'^-ixy+y'^-4x+2y=0. 

3.  x'^-^2xy-Sy^-6x+6y=0,  xy  +  y-2=0. 

4.  x^+xy-Sy^-Sx+4y-\-9=0,  xy-2z=0. 

6.  2x2-4a:y-j/2_8x+8y+33=0,  a;2-4aj+i/2-21=0. 

6.  x'^-j-xy  +  y'^+x-\-y-S=0,  xy-{-2x+2y-^i=0. 

7.  x^-\-xy-~y^+ax—ay+c=0,  xy  —  ax+ay+c=0. 

8.  x^+Sxy+2y^+x+y=0,  a;«+y«+(a;+y)(a;-2)(2/-l)=0. 

9.  2x^+xy-4x-l=0,  y'^-^2xy-^y+l=0. 

10.  x?-[-y^-\-l-Sxy=0,  xy+6=0. 

11.  x^+y^+z^—Sxyz=0,  ax-\-by-{-cz=l,  lx+my+nz=0. 

12.  Solve  x^  +  px^  +  gx2  4-  rx  4-  s  =  0  by  assuming  x2  =  y,  and 
assigning  to  X  a  value  that  will  make 

y2  +  pxy  -\.  qy  ^  rx  +  s  +  \{x^  -  y) 
break  up  into  linear  factors  in  x  and  y. 


598  horner's  method. 


?  +  ^>0; 


CHAPTER  XLI. 

horner's  method. 

415.  The  cube  roots  of  2/^  and  z^  in  Cardan's  formula  can- 
not be  found  by  the  arithmetical  cube-root  process  unless 

4   '  27 

and  even  when  this  condition  is  satisfied  the  various  re- 
ductions are  frequently  quite  laborious.  In  the  solution 
of  numerical  equations  a  process  of  direct  numerical 
approximation  is  found  to  be  more  expeditious.  Horner's 
Method,  first  published  in  1819,  based  on  Horner's  process 
of  synthetic  division  [Art.  83],  is  applicable  to  finding 
the  real  roots  of  equations  of  any  degree. 

As  a  preliminary  step  in  establishing  Horner's  method 
we  require  the  proposition  of  the  next  article. 

416.  To  transform  a  multinomial  in  x  into  a  multi- 
nomial in  (x  —  a).  J" 

For  our  present  purpose  it  will  be  sufficient  to  apply 
this  transformation  to  a  multinomial  of  the  fourth 
degree.^* 

When  the  binomials  (a  +  ^)*,  (a-\-hy,  {a  +  lif,  in  the 
expression 

M=za{a  +  hy-\-p{a-\-  lif  +  g(a  -t-  hf  +  r (a  H-  /i)  +  s, 

*  See  Art.  446  for  a  discussion  of  the  general  case. 


hoener's  method.  599 

are  expanded,  and  the  coefficients  of  the  several  powers 
of  h  are  collected,  the  result  must  be  of  the  form 

in  which  ijg,  R^,  i?i,  Rq  involve  a  but  not  h.     li  x  —  a 

replace  h  in  these  two  equivalents  of  Jf,  the  result  Is  the 

identity 

ilif  =  aic*  4-  pic^  -\- qx^  -\- rx -{-  s 

^a{x-ay  -\-  Rz{x-ay  +  Ri{x-af  -\-  R^{x-a)  -\-  Rq. 

Let  M  be  divided  by  a;  —  a,  giving  as  quotient  Mi ;  the 
remainder  is  Rq  and 

Jfi  =  a (ic  -  af  -\-R^{x-  af  +  i?2 (a;  -  a)  +  R^j 
Jf=  Jfi(a;-a)  +  i?o. 

Similarly,  let  Jfs,  J^fg,  Jf4  denote  the  successive  quotients 
of  Ml  hy  x—a,  M2  by  ic— a,  and  M3  by  a;— a,  respectively ; 
then 

Mi  =  M2{x  —  a)  -f  i?i, 
M2  =  Ms(x  —  a)  -\-  R^f 
Ms  =  J/4  (a;  -  a)  +  i?3, 
3/4  =  a. 

Thus  the  coefficients  Rq,  Ri,  R^,  R3,  of  the  transformed 
multinomial,  are  the  successive  remainders  in  these 
several  divisions,  and  a  is  the  last  quotient,  namely,  the 
quotient  which  does  not  contain  x  —  a. 

It  is  obvious  that,  although  the  multinomial  is  here 
of  the  fourth  degree,  this  process  of  transformation  is 
applicable  to  multinomials  of  any  degree,  and  that  the 
rule  may  be  stated  in  the  following  general  terms : 

Divide  the  given  multinomial  by  x  —  a,  and  set  aside  the 
remainder  as  the  teryi  not  containing  x  —  a.     Divide  the 


600  HORNER'S  METHOD. 

quotient  of  this  division  by  x  —  a,  and  set  aside  the  remain- 
der as  the  coefficient  of  x  —  a.  Repeat  this  process  until 
there  is  a  quotient  which  does  not  contain  x  —  a.  This  last 
quotient  will  be  the  coefficient  of  the  highest  power  of  x—a. 

Ex.  1.  Transform  x^  —  11  x^ -\-  15  x  -\-  27  into  a  multinomial  in 
(x  -  9). 

Divide  the  given  multinomial  by  x  —  9,  using  detached  co- 
efiBcients.     It  is  convenient  in  this  work  to  place  the  divisor  at  the 

"^^*'  1        -11  15  27        [9=  a 

9        -18        -^ 

-  2        -  3  0  =  Bo 

_9  63 

7  60  =  i^i 

_9 
1  16  =  i?2 

Thus,  the  transformed  multinomial  is 

(X  -  9)3  +  16(a;  -  9)2  +  60(a;  -  9); 
and  it  is  now  obvious  that  9  is  a  root  of  the  equation 
a;3  _  11  a;2  _|.  15  a;  +  27  =  0. 
Ex.  2.   By  trial  it  is  found  that  the  equation 
4ic8-9x2-14a;  +  4  =  0 

has  a  root  between  .2  and  .3  ;  find  this  root. 

Transform  Ax^  —  9 x^  —  14t  x  +  4:  into  a  multinomial  in  (x  —  .2). 

4        ^9.0  -14.00  4.000        [^ 

^  -  1.64        -3.128 

-8,2  -15.64  .872  =  Bq 

'    \     ^  -  1.48 

**      *'    *_7.4  -17.12  =  i?i 
.8 


4         -6.6  =  B2 
The  transformed  multinomial  is 

4  y8  -  6.6  2/2  _  17.12  y  +  .8J2^  r^  x 


^. 


Horner's  method. 


601 


For  a  trial  value  of  y,  neglect  the  first  two  terms  and  write 

-  17.12  y  +  .872  =  0  ;  y  =  .05. 
Divide  4  y'  -  6.6  y'^  -  17.12  y  +  .872  hj  y  -  .05. 


-6.6 

-17.12 

.872 

.2 
-6.4 

-     .32 
-17.44 

-.872 
0 

.2 
-6.2 

-     .31 
-17.75  =  i?i 

.2 

1:25 


0  =  i?o 


4  -6    =i?2 

The  transformed  multinomial  is 

4  ;s8  _  6  ^2  _  17.75  z;   z  =  y-..05  =  x-  .25. 

The  equation  in  x  can  now  be  written  in  the  form 

4Cx  -  .25)8  _  6(x  -  .25)2  _  i7.75(a;  _  .25)  =  0, 

of  which  .26  is  at  once  seen  to  be  a  root. 

The  two  parts  of  this  work  can  be  placed  together,  and  the 
whole  compactly  arranged  as  follows  : 

|.26  =  tt 


^0 


-9.0            -14.00 
.8            -  1.64 

4.000- 
-3.128 

-8.2 
.8 

-15.64 
-  1.48 

.872 
-  .872 

-7.4 

.8 

-17.12 
-     .32 

0 

-6.6 
.2 

-6.4 
.2 

-6.2 
.2 

-17.44 
.31 

-17.75  = 

I 

h 

-6 


J?2 


For  the  completed  value  of  a  to  be  an  exact  root,  it  is 
necessary  and  sufficient  that  Rq  =  0,  and  the  smaller  Hq 
is,  the  nearer  does  a  approximate  to  a  root. 


602 


HORNER  S   METHOD. 


Ex.  3.   The  equation 

X*  -  16  x3  +  64  x2  -  60  =  0 

has  a  root  between  1  and  2.    Find  an  approximate  value  for  this 
root. 


-16       64         0 
1      -15        49 

-60      11.127 
49 

-15       49 
1      -14 

49 
35 

-11.0000 
8.6081 

-14 

1 

35 
-13 

84.000 
'2.081 

-  2.39190000 
1.76817136 

-13 
1 

22.00 
-  1.19 

t 

86.081 
1.963 

-  .623728640000 
.622246188641 

1  -12.0 
.1 

20.81 
-  1.18 

88.044000 
.364568 

-  .001482451359 

■ 

-11.9 
.1 

19.63 
-  1.17 

88.408568 
.359944 

-11.8 
.1 

18.4600 
-  .2316 

88.76851200 
.12380066 

0 
3 

-11.7 
.1 

18.2284 
-  .2312 

88.892312663 
.123236869 

1-11.60 
.02 

17.9972 
-  .2308 

89.01554953201 

)0 

-11.58 
.02 

17.766400 
-  .080591 

-11.56 
.02 

17.685809 
-  .080542 

-11.54 
.02 

17.605267 
-  .080493 

1-11.520 
.007 

17.52477400 

1 

-11.513 
.007 

-11.506 
.007 

-11.499 
.007 

I  -11.4920 

The  next  figure  in  this  root  is  easily  found  to  be  zero. 


HORNEK  S   METHOD. 


603 


Ex.  4.    Find  the  cube  root  of  5. 
The  equation  to  be  solved  is 
efficients  of  x^  and  x  are  zero. 


x3  —  5  =  0.     Note  that  the  co- 
The  process  of  solution  is  as 


follows : 


5 


11.71 


4.000 
3.913 


2 

1 

3.00 
2.59 

-  0.087000 
87211 

3.0 

7 

5.59 
3.08 

+  .000211 

3.7 

7 
4.4 

7 

8.6700 
511 

8.7211 
512 

5.10 

1 

8.7723 

6.11 

1 

6.12 

1 

5.13 


This  result  is  too  large  by  .000211  ;  that  is  (1.71)8  =  5  +  .000211. 
Expressed  in  another  way,  .000211  is  the  result  of  substituting  1.71 
for  a:  in  xS- 5.     [Art.  149.] 

For  methods  of  abbreviating  Horner's  method  the  student  is 
referred  to  the  larger  works  on  the  theory  of  equations. 


EXAMPLES  cm. 

Apply  Homer's  method  to  finding,  either  exactly  or  approxi- 
mately, the  real  roots  of  the  following  equations  : 


1.  a:2_4aj-}-2 

2.  a;8-2  =  0. 


0. 


3.  x3-6a;-6  =  0. 

4.  2  «8  _  7  a;2  +  2  X  +  6  =  0. 


604  HORNER'S   METHOD. 

5.  4a;3- 18a;2  +  23x- 7.5  =  0. 

6.  2x8-85x2-85x-87  =  0. 

7.  CK*  -  a;3  -  2  a:2  +  3  cc  -  3  =  0. 

8.  o^-x^  +  x^-Sx^  +  Sx-S=0.    (Resolve  into  factors.) 

9.  x8  -  3  a;3  -  2  a;5  +  6  =  0.     (Resolve  into  factors.) 
10.  x3  +  12x-12  =  0. 


BOOTS  AND  COEFFICIENTS.  605 


CHAPTER  XLII. 

Roots  and  Coefficients.     Symmetric 
Functions. 

417.  The  general  rational  integral  function  of  a;,  of 
degree  n,  is 

f{x)  =  ttoa;"  +  CLi^'^  H h  a^-iX  +  «„• 

It  is  sometimes  desirable  to  discuss  it  under  the  form 

P(X)  =  «"  +i)ia;"-l  ^ [- Pn-l^ -\- Pny 

in  which  the  coefficient  of  a;**  is  unity.     The  two  forms 
are  directly  connected  with  one  another  by  writing 

Then  f(x)  =  aoP(x), 

and  the  roots  of  the  two  equations, 

f(x)  =  0,  P(x)  =  0, 
are  the  same. 

In  the  present  discussion  it  is  assumed  that  there 
exists  at  least  one  number,  real,  imaginary,  or  complex, 
which,  when  it  is  substituted  for  x  in  P(a;),  will  make 
P(x)  =  0.*  Such  a  number  is  called  a  root  of  the  equa- 
tion P(x)  =  0  [Art.  327]. 

*  See  the  larger  works  on  the  Theory  of  Equations,  for  a  proof 
of  this  theorem. 


606  EOOTS   AND   COEFFICIENTS. 

Under  this  assumption  it  follows  [Art.  325]  that  any 
multinomial  P(x)  of  the  'n}^  degree  has,  as  one  of  its 
forms, 

F(X)  =  (X-  aXx  -  I3)(X  -y)...(x-  v), 

where  a,  ft  y,  •••  v,  are  the  n  roots  of  P(x)  =  0. 
.  If  we  wish  to  express  f(x)  in  this  form,  we  write 

f(x)  =  af,{x  -  a)(x  -/3)'..(x  —  v). 

This  product-form  of  a  multinomial  calls  to  our  atten- 
tion the  important  fact,  that  the  problem  of  finding  the 
roots  of  an  equation  f(x)  =  0  is  essentially  identical 
with  that  of  resolving  the  multinomial  f(x)  into  its 
linear  *  binomial  factors.  Since  there  are  just  n  of  these 
factors  [Art.  325],  an  equation  of  the  n^^  degree  has  just 
n  roots. 

418.  By  expanding  the  product-form  of  f(x)  or  P(x)j 
we  can  obviously  express  the  coefficients  a^,  aj,  ••-,  or 
Pif  P2>  ••*?  as  functions  of  the  roots  a,  ft  y,  •••  v.  Thus 
the  expanded  form  of  (x  —  a){x  —  /5)(a;  —  y)  is 

a^  -  (a  -f  )8  +  y)  0^  +  (/8y  +  ya  -f-  a/3)  a;  -  a^y. 

Hence,^  if  the  cubic  equation  be 

a^-\-px^  +  qx  +  r=Of 

and  a,  ft  y,  be  its  roots,  the  following  relations  exist 
between  the  coefficients  and  the  roots : 

2)  =  -  (a  -f  ^  4-  y),  5'  =  iSy  +  ya  -f  aft  r  =  —  aj8y. 

*  First  degree  factors.  ^ 


ROOTS  AND  COEFFICIENTS.  607 

Passing  to  the  equation  of  the  n*^  degree,  P(x)  =  0, 
and  denoting  its  roots  by  a,  y8,  y,  •••  v,  and  writing 

Sa  =  a  +  )8  +  y  +  SH \-v 

=  sum  of  the  roots, 
^j8  =  a/3  +  ay  +  aS  H h  av  +  )8y  +  /SS  +  ••• 

=  sum  of  products  of  pairs  of  roots, 
Sa^y  =  a)8y  +  a^8  +  .••  +  a^v  +  ayS  +  — 
=  sum  of  products  of  roots  in  threes, 

a^y ...  V  =  product  of  all  the  roots, 
the  expanded  form  of  (a;  —  a) (a?  —  P)(x  —  y)  -"  (x  —  v)  is 

05"  -  JC'-^Sa  +  iC"-22ay8  +  a;«-32a^y  +  .-  +  a^y ...  v; 

and  in  order  that  this  may  be  identical  with 

af*  +i>ia5"-^  +p^-^  +P3«""^  H \-Pn, 

there  must  exist  between  pi,  p2f  p^,  ••- Pnt  and  a,  ^,  y,  ...  v, 
the  following  relations  [Art.  328] : 

Pi  =  -  2a,  Pa  =  Saft  Pg  =  -  ^^7,    —  2>«  =  ±  o.Py  —  v. 

r/iese  equations  state  the  simple  law  that  connects  the 
coefficients  and  the  roots  of  the  general  equation  of  the 
n^  degree. 

Since  there  are  n  of  these  equations,  and  just  n  roots, 
the  enquiry  naturally  arises:  why  is  it  not  possible  to 
find  a,  ft  y,  •••  V,  by  some  process  of  elimination?  The 
answer  is,  that  any  such  process  necessarily  leads  back 
to  the  equation  of  the  n*^  degree  with  which  our  investi- 
gations began.     Nevertheless,  the  knowledge  of  these 


608  BOOTS  AND  COEFFICIENTS.      ^ 

relations  between  roots  and  coefficients  is  an  important 
aid  in  the  solution  of  equations. 

419.  Symmetric  Functions.  The  expressions  %a,  %a^, 
^aySy,  etc.,  are  evidently  such  that  the  interchange  of  any 
two  of  the  letters  leaves  the  expressions  themselves  un- 
changed ;  they  are  therefore  symmetric  functions  of  the 
roots.  [Art.  150.]  Such  functions  are  important  in 
the  general  theory  of  higher  equations. 

Any  symmetric  function  of  the  roots  can  be  expressed 
in  terms  of  the  coefficients  of  the  given  equation.  A 
formal  proof  of  this  statement  is  not  presented  here; 
a  few  examples,  as  verifications  of  particular  cases,  must 
suffice  for  this  elementary  presentation. 

Ex.  1.  Sa2  =  (Sa)2  -  2  SajS  ^ 

Ex.  2.  Sa8  =  (Sa)8  -  3  Sa2j3  -  6  Sai57. 

Also  Sa8  =  Sa2  .  Sa  -  Sa2/3. 

/.  2  Sa3  =  3  Sa2  .  Sa  -  (Sa)3  +  6  Sa/Sy 
=  -  3i)i(i>i2  -  2i)2)  +  Pi^  -  eps 
=  -  2pi3  +  6piP2  -  Qps. 
,\  Sa3  =  _  pi8  +  Spip2  -  3p$.      ' 

Ex.  3.  Sa2^  =  Sa2  .  Sa  -  Sa^ 

=  -  Pi(Pi^  -  2i)2)  +  i?!^  -  3i)ii?2  +  3i)8 

=  -PlP2+Sp3. 

Ex.  4.  To  solve  the  cubic  equation  by  the  aid  of  symmetric 
functions. 


BOOTS   AND   COEFFICIENTS.  609 

Denote  by  a,  /3,  7,  the  roots  of  the  general  cubic  equation 

x^  +  px"^  -i-  qx  +  r  =  0, 

and  take  note  of  the  relations 

a  +  /3  +  7  =  ~p, 

/Sy  +  7a  +  a/3  =  g, 

a/37=-r.  [Art.  418.] 

Then,  denoting  by  1,  w,  w2^  i^^q  three  cube  roots  of  1,  and  by 
m  and  n,  the  trinomials  a  +  w/3  +  u^  and  a  +  w^^  +  ory  respec- 
tively, we  have  three  linear  equations 

a  +  /3  +  7=-p, 

a  +  «/3  +  u^  =  niy 

a  +  w2/3  +  ary  =  w, 

from  which,  with  the  help  of  the  relations  1  +  w  +  w^  =  q,  w^  =  1, 
we  obtain,  by  easy  eliminations, 

a  =  i(-P  +  w  +  n), 
/S  =  \(—p  +  w^m  +  wn), 
7  =  ^(  — p  +  wm  +  w^n). 

The  problem  of  solving  the  cubic  is  now  reduced  to  the  finding 
of  m  and  n. 

The  product  of  the  two  expressions  for  m  and  n  gives  [Art.  406] 

rnn  =  a2  +  /32  -f  72  -  (/37  +  7a  +  ap) 
=  (a  4-  /3  +  7)2  _  3(/37  +  ya  +  ajS) 
=  p2_3g; 

and  the  product  of  the  three  expressions  for  a,  /3,  7  is  [Art.  406] 

apy  =  —  r  =  ^^  (  —  p^  +  m^  +  ?i*  +  Sjpmn) ; 

and  from  the  last  equation  we  find 

w8  +  w8  =  p3  _  3p(p2  _  3  ^)  _  27  r 

=  -2i?8  +  0pg-27r. 
2s 


610  ROOTS   AND   COEFFICIENTS. 

For  the  sake  of  brevity  let 

/,  gf  =  -  2p3  +  ^pq  -27r,  p^-Sq; 

then  m^  -{■  n^  =f,  mn  —  g^ 

f  and  g  being  known  quantities,  and  the  solution  of  these  equa- 
tions in  m  and  n  gives  immediately 


The  real  cube  roots  of  these  two  quantities  (numbers)  can  be 
found  as  soon  as  the  numerical  values  of  /  and  g  are  given,  pro- 
vided 

and  the  roots  of  the  cubic  are  then  obtained  in  the  form 

—  p  +  m  +  n     —  p  -\-  ttfVt  4-  (tin     —  p  +  (^m  +  to^w 
3  '  3  '  3  * 

But  if  /2  <  4  g^,  real  cube  roots  of 


Kf+^f'-^9^)    and    i(f-Vrz-4g^)    . 

cannot  be  obtained  by  the  usual  arithmetical  process.     The  follow- 
ing trigonometrical  method  is  then  effective. 
Write,  for  the  moment, 


h=y/ig^-f^  so  that   Vp  -  4  g»  =  ih, 
where  i  stands  for  V—  1  and  h  is  real.     Then 

and  since  (^)%  (-^V  =  ^:  =  1, 

we  may  write        —^—=  =  cos  3  d,   — — =  =  sin  3  d, 
2V^  2V^3 


ROOTS    AND   COEFFICIENTS.  611 

The  substitution  of  cos  3  0  and  sin  3  6  for  their  equivalents  in  the 
expressions  for  m^  and  #  now  gives 

m3  =  V^(cos  3  ^  +  I  sin  3  ^), 

n3  =  Vg^(cos  Sd-isinSe), 

and  by  the  taking  of  cube  roots 

m  =  Vg(cos  6  +  i  sin  ^), 

n  =  Vg(co8  e  —  i  sin  6). 

EXAMPLES  CIV. 

Verify  the  following  relations.  Take  note  that  they  have  refer- 
ence to  an  equation  of  the  w'*^  degree  having  n  literal  coefl&cients 
Ph  P2»  •••»  Pm  and  n  roots  o,  /3,  •••,  p,  and  that 

pi  =  -  Sa,  p2  =  SajS,  ps=-  2a/37,  ^4  =  Sa/375. 

1.  pi2  =  Sa2  +  2  SajS. 

2.  P18  =  -  Sa*  -  3  Sa2^  -  6  Sa/37. 

3.  pi;)2  =  -  Sa2/3  -  3  So/37. 

4.  pi*  =  Sa*  +  4  Sa^iS  +  6  Sa2/32  +  12  Sa2/37  +  24  Sa/S78. 

5.  PiPs  =  Sa2/37  +  4  Sa/375. 

6.  ^2^  =  Sa2/32  +  2  Sa2/37  +  6  Sa/975. 

7.  I)i2p2  =  2a8/3  +  2  Sa2/32  +  5  Sa2/37  +  12  Sa/378. 

8.  Sa*  =  pi*  —  4ipi^p2  -\-2p2^  +  4pip3  -  ip^. 

9.  Sa3/3  =  pi2p2  -  2  p2^  -  pip3  +  4  Pi- 
le.   Sa2j32  =  p2^  -  2piP3  +  2  Pi- 
ll.   Sa2/37  =piP8  —  4p4. 

12.    Sa8/375  =pi2p4-2p2P4. 

/  13.     Sa2^2^2=^g2_2p2p4. 


^12  KOOTS   AND   COEFFICIENTS. 

14.   Denote  by  a,  ^,  7,  5  the  four  roots  of 

x^ -{- px^  +  qx"^ -i- rx  +  s  =  0, 
let  A  =  l(ap  +  yd),  B  =  i(a7  +  )85),    C=  l(ad  +  ^37), 

and  prove  the  following  relations  : 

A-j-B  +  C=lq, 
BC  +  CA  +  AB  =  l(pr  -  4  s), 

ABC=i(r^+p2s-iqs). 

15.   Verify  that  the  values  2A,2B,2C,  of  Ex.  14,  are  the  roots 
of  the  reducing  cubic  of  Art.  410. 


GRAPHICAL  METHODS. 


613 


CHAPTER   XLIII. 

Graphical  Methods. 

420.   Graphs.     If  f(x)  be  any  rational  integral  function 
of  Xj  and  it  be  assumed  that 

then  to  every  value  of  x  there  corresponds  one  value  of 
y.  Let  the  corresponding  values  of  x  atod  y  be  repre- 
sented by  lengths  laid  off  on  two  fixed  straight  lines 
which  intersect  at  right  angles  in  a  point  0,  the  values 
of  X  on  the  horizontal  line  OX,  the  values  of  y  on  the 
vertical  line  0  Y.  Lines  extending  from  0  F  to  the  right 
shall  be  regarded  as  positive,  those  extending  to  the  left 
negative ;  and  lines  extending  from  OX  upward  shall  be 
regarded  as  positive,  those  extending  downward  negative ; 
and  these  positive  and  negative 
lengths  shall  correspond  respec- 
tively to  positive  and  negative 
values  of  x  and  y. 

Suppose  OM,  ON  (Fig.  1)  to 
represent  a  pair  of  values  of  x, 
y ;  through  M,  N  draw  straight 
lines  MP,  NP,  parallel  respec- 
tively to  OY,  OX,  and  intersecting  in  P.  To  this  pair 
of  values  of  x,  y  there  now  corresponds  the  point  P,  and 


Y 

N 

y 

p 

0 

X       ^ 

/I            X 

Fig.  1. 


614  GRAPHICAL  METHODS. 

to  every  pair  of  values  x,  y  there  corresponds  one,  and 
only  one,  such,  point.  It  is  customary  to  designate  it  as 
the  point  (x,  y). 

If  X,  y  be  now  supposed  to  undergo  simultaneous  and 
continuous  change,  but  always  in  such  a  way  as  to 
satisfy  the  equation 

the  point  P  will  move  in  its  plane  and  trace  out  a  line, 
in  general  a  curved  line,  or  curve.  This  curve  is  called 
the  graph  of  the  function  f(x),  and  y  =f(x)  is  said  to  be 
its  equation.  Every  rational  integral  function  has  such 
a  graph. 

The  nomenclature  of  analytic  geometry,  as  applied 
to  this  graphical  representation,  is  convenient  and  is 
as  follows :  x,  y  are  the  rectangular  co-ordinates  of 
the  point  (x,  y),  x  its  abscissa,  y  its  ordinate;  OX, 
OY  are  the  co-ordinate  axes,  and  0  is  the  origin  of 
co-ordinates. 

Other  forms  of  graphical  representation  may  be  em- 
ployed, but  the  foregoing  is  the  most  useful  for  our 
present  purposes. 

421.  Given  any  function  f{x),  an  indefinite  number  of 
isolated  points  on  its  graph  may  be  determined,  through 
which  the  curve  may  be  roughly  drawn  free-hand.  First 
assign  a  succession  of  small  positive  and  negative  values 
to  X  (the  exaniples  here  given  will  not  require  large 
values),  then  compute  the  corresponding  values  of  f{x). 
A  few  examples  will  illustrate  the  process. 


GRAPHICAL  METHODS. 


615 


Ex.  1.   Construct  the  graph  of  x"^. 

The  equation  y  =  x^  yields  the  following  series  of  corresponding 
values  of  x  and  y,  tabulated  in  vertical  columns : 


(-3,9) 


y  = 

=  aj2 

X 

y 

-3 

9 

-2 

4 

-1 

1 

0 

0 

1 

1 

2 

4 

3 

9 

These  seven  points  (0, 0),  (1, 1),  (-1, 1),  (2, 4),  (-2,  4),  (3,  9), 
(  —  3,  9)  serve  to  determine  the  general  trend  of  the  curve,  which  is 
as  represented  in  Fig.  2.    Other  points  can  be  easily  interpolated. 

Ex.  2.   Construct  the  graph  of  x"^  -2x-  1. 
The  equation  y  =  x^  —  2x  —  I  yields  the  following  system  of 
corresponding  values,  and  the  graph  is  represented  in  Fig.  3. 


y  =  x^-2x-l 


e2,7)\ 

/a7) 

(-1.2)\ 

/ 

K3.2) 

-    \ 

(0,-1) 

0 

/ 

X 

-13 

(1,- 

-3) 

X 

y 

-3 

14 

-2 

7 

-1 

2 

0 

-1 

1 

-2 

2 

-1 

3 

2 

4 

7 

5 

14 

Fig.  8. 


616 


GRAPHICAL   METHODS. 


It  is  obvious  from  an  inspection  of  the  figure  that  there  is  a 
value  of  X  between  —  1  and  0,  and  another  between  2  and  3, 
which  will  make  a;^  —  2  cc  —  1  vanish.  Hence  there  is  one  root  of 
the  equation  x^  —  2x  —  1=0  between  —  1  and  0  and  another  be- 
tween 2  and  3. 

The  curves  of  Figs.  2  and  3  have  axial  symmetry  with  respect 
to  a  straight  line.  The  axis  of  symmetry  in  Fig.  2  is  OF,  in  Fig.  3 
it  is  a  line  parallel  to  OF  at  unit's  distance  to  the  right  of  the 
origin.  The  graph  of  every  function  of  the  form  ax^  +  fta:  +  c  has 
such  an  axis  of  symmetry. 

Ex.  3.   Construct  the  graph  of  x^  —  9  a:. 

In  order  that  the  values  of  y  may  be  conveniently  small  we 
make  y  =  \{x^  —  9x). 


y^\{x?-^x). 


(-2,5y> 

y 

\ 

U 

A 

0 

^(3,-5) 

X 

y 

-4 

-14 

-3 

0 

-2 

5 

-1 

4 

0 

0 

1 

-   4 

2 

-    5 

3 

0 

4 

14 

Fig.  4. 


This  curve  has  symmetry  With  respect  to  the  origin, 
points  are  represented  in  the  figure  at  3,  —  3  and  0. 


The  root 


It  will  be  observed  that  root-points,  that  is,  points  whose 
abscissae  are  roots  of  the  equation  f(x)  =  0,  lie  at  the 
Intersection  of  the  graph  and  the  a;-axis. 


GRAPHICAL  METHODS.  617 

EXAMPLES  CV. 
Construct  the  graphs  of  the  following  functions  : 

1.  2x-l.  4.    ic2-2jc-4.  T.   x^-7^-\-x-\. 

2.  (x-l)2.  5.    x^-2x.  8.    X*- 6x^  +  6. 

3.  ^2  -  X  +  1.  6.    a;3  -  2a;  +  1.  9.   x^  -  3x2  _  6a;. 
Note.  —  The  graph  of  example  9  consists  of  two  distinct  parts. 

BINOMIAL  EQUATIONS. 

422.  In  the  foregoing  graphical  scheme  only  real  roots 
are  represented  in  the  diagram.  In  general,  for  the 
graphical  construction  of  complex  roots  a  plane  is  nec- 
essary, and  the  analysis  for  the  general  case  is  very 
complicated.  But  the  root-points  of  so-called  binomial 
equations,  whose  form  is 

aaj'*  -f  6  =  0, 

may  be  constructed  as  points  on  the  circumference  of  a 
circle,  whose  radius  is  the  positive  numerical  ?i*^  root 
of  ±  b/a,  of  +  b/a  if  b/a  be  negative,  of  —  b/a  if  b/a  be 
positive. 

Let  a  be  this  numerical  n^^  root;   the  equation  then 

has  the  form 

a;"  ±  a**  =  0, 

and  there  are  two  distinct  cases,  a;"  =  a",  and  x^  =  —  a**, 
which  present  two  corresponding  types  of  solution : 

(1)  X  =  a  Vl ;      (2)  X  =  a^-^. 

Since  an  equation  of  the  n^^  degree  has  n  roots,  there 
are  n  values  of  Vl,  as  also  of  -\/—l,  which  are  the  n 
roots  of  2;"  =  1,  or  of  2;"  =  —  1.     These  roots  having  been 


618 


GRAPHICAL  METHODS. 


Fig.  5. 


found,  the  n  roots  of  x''  =  a""  or  of  x""  =  —  a"  are  the  n 
values  of  aVl,  or  of  aV—  1.  The  problem  of  solving 
ax''  +  6  =  0  is  thus  reduced  to  that  of  solving  equations 
of  the  form  2"  =  ±  1. 

423.  A  complex  quantity  x-{-iy  (i  =  V— 1)  contains 
two  numbers,  x,  y,  which  may  be  regarded  as  the  rec- 
tangular co-ordinates  of  a  point 
.  (Fig.  5).  Thus  interpreted,  every 
complex  number  has  assigned  to 
it  a  point,  its  graph,  in  a  plane, 
and  to  every  point  there  corre- 
sponds a  complex  number,  its 
affix.  The  points  we  are  about 
to  consider  lie  upon  the  circumference  of  a  circle  of  unit 
radius,  a  U7iit  circle,  whose  centre  is  at  the  origin ;  their 
co-ordinates,  therefore,  satisfy  the  condition 

'X^  +  f  =  l. 

\ 
If  two  complex  numbers  x  +  iy,  x'  -\-  iy',  whose  parts 

X,  y,  x',  y\  satisfy  this  condition,  be  multiplied  together, 

the  result  will  be  of  the  form  u  +  iv,  that  is, 

{x-\-  iy)(x'  +  iy')  =  u-{-  iv, 

where  u  =  xx'  —  yy',   v  =  xy'  -f  x'y, 

and  it  can  be  verified  at  once  that 

u^^v^^ix'-h  y")  (x'^  +  y"^  =  1. 

Hence,  if  the  graphs  of  a  series  of  complex  numbers  lie 
upon  the  unit  circle,  so  does  also  the  graph  of  their  product. 


GKAPHICAL  METHODS. 


619 


If  there  be  n  equal  factors,  the  product  has  the  special 
form  {x  +  lyY  and  hence,  in  particular. 

If  {x-\-  iyY  =  u  -{-iv  and  oi?  -\-]f  —  1,  then  w^  +  i;^  =  1 
and  the  graph  of  u-\-  iv  is  on  the  unit  circle. 

424.  Given  x-\-iy  and  a?'  +  iy\  it  is  required  to  con- 
struct the  product  (x  +  iy){x'  -\-  iy'). 

Let  OM,  MP=x,y, 
OR,  BQ  =  x',  y\ 

(Fig.  6)  and  let  u,  v,  be  the  hor- 
izontal and  vertical  co-ordinates 
of  Q,  so  that 

ON,  NQ  =  u,  V. 

Then,  in  the  similar  triangles 
QSR,  OMP,  and  SON,  POM, 

QS  :  y'  =zl:x,  SN:  u  =  y  :  x, 
1 


Fig.  6. 


and 


v=:qS  +  SN=^l{y^  +  uy). 

X 

.'.  y'  =  vx  —  uy. 


Also, 


x'^=l-y'^  =  {u^-\-v^{7?-\-f)-iyx-uyy 

=  (ux  -f  vyy. 
.'.  a;'  =  wa;-f  vy. 
In  the  figure  here  constructed  it  is  obvious  that  the 
plus  sign  for  this  value  of  a'  must  be  chosen.     It  can  be 
shown  that  this  choice  is  necessary  in  all  cases. 

By  forming   the   expressions    xx^  —  yy',  x'y  -f-  xy',  we 

now  have 

xx'  —  yy'  =  ux^  -\-  uy^  =  u, 

xy'  -f  x'y  =  vy^  -(-  va^  =  v. 


620  GRAPHICAL  METHODS. 

But        {x  +  iy)(x'  +  iy^)  =  xx'  —  yy'  -\- i (xy'  +  x^y). 

.'.  (a?  +  iy)(x'  +  iy')  =  w  +  iv, 

which  identifies  Q  as  the  required  graph  of  the  given 
product. 

Thus  it  has  been  shown  that  the  graph  of  the  product 
(x  +  iy)  (x'  +  iy')  is  constructed  by  first  determining  P 
as  the  graph  ofx-\-  iy,  in  the  usual  way,  and  then  con- 
structing the  point  (x',  y'),  using  OP,  instead  of  OA,  as 
the  axis  of  abscissae.  By  repeating  this  process  the 
product  of  any  number  of  factors  may  be  constructed. 

It  is  now  evident  that  the  graph  of  the  series  of  consecu- 
tive powers, 

(x  +  iyj,  {x  +  iy)\  (x  +  iyf,  ..•(«;  +  lyy, 

is  a  series  of  points  distributed  at  equal  intervals  over  the 
circumference  of  the  unit  circle. 

For,  the  constructions  for  these  powers  present  a 
series  of  congruent  right  triangles  OMP,  ORQ,  etc.,  and 
the  arcs  AP,  PQ,  etc.,  are  therefore  equal. 

425.  Any  root-point  of  the  equation  z""  =  1,  or  of  the 
equation  z^  =  —  1,  is  on  the  unit  circle. 

For,  let  to  be  the  root  in  question,  and  let  {x,  y)  be  the 
corresponding  root-point.     Then 
o)  =  a?  -f  I?/ 

where  x'  =  xj-yjiy?  +  2/^), 

and  we  may  write 

where         u-\-iv  =  (x'  +  iy')\ 


GRAPHICAL  METHODS.  621 

But  this  demands  that  u^-\-v^=l  [Art.  423]  and  that 
v  =  0,  u  =  ±l  [Art.  234]. 

426.  If  0)  be  a  root  of  z'^  =  1,  any  integral  power  of  w 
is  also  a  root;  for  then,  k  being  any  integer, 

(a>*)"  =  (a>»)*  =  1*  =  1. 

Ifoibea  root  o/  2**  =  —  1,  any  odd  integral  power  of  w 
is  also  a  root;  for  then,  k  being  any  odd  integer, 

(o>*)'*  =  (a)")*  =  (-l)*  =  -l. 

Ifo/'  =  l,  and  ifti/"  be  the  lowest  power  ofo)  that  has  this 
value,  0),  0)^,  o>^,  •  •  •  (!)"■,  are  all  distinct,  and  are  the  n  roots 
of  z^  —  1.  For,  if  any  two  are  equal,  as  w^,  o>*,  where  I 
and  k  are  less  than  n,  then  a>'~*  =  1,  which  contradicts 
the  hypothesis. 

7/"  to**  =  —  1,  and  ifw'*  be  the  lowest  power  of  o)  that  has 
this  value,  w,  w^,  o>*,  •••w'^""^,  ai-e  the  n  roots  0/2**  =  — 1. 
For,  tu^"  =  +  Ij  and  this  is  the  lowest  power  of  a>  that 
has  the  value  4- 1 ;  thus  o,  <o^,  w^,  •••  o)^",  are  distinct,  and 
therefore  also  w,  w^,  <o^,  •••  a>^**~\ 

427.  The  root-points  of  2'*=  ±  1  may  now  be  definitely 
placed  on  the  circumference  of  the  unit  circle. 

I.  When  2"  =  + 1.  For  all  values  of  n,  1  is  a  root. 
Beginning  with  (1,  0)  as  a  first  root-point,  the  whole 
series  of  n  root-points  are  distributed  at  equal  intervals 
over  the  entire  circumference. 

II.  When  2"  =  —  1.  For  no  values  of  n  is  1  a  root. 
If  2n  points,  of  which  (1,  0)  is  one,  be  distributed  at 
equal  intervals  over  the  circumference  and  be  numbered 


622 


GRAPHICAL  METHODS. 


consecutively  1,  2,  3,  4,  •••,  2w,  and  be  so  adjusted  that 
2  71  is  assigned  to  the  point  (1,  0),  the  odd  numbers  will 
mark  the  positions  of  the  n  root-points. 

Ex.  1.    Solve  the  equation  z^  =  —\. 

Divide  the  circumference  into  six  equal  parts  and  number  the 
division-points  1,  2,  3,  4,  5,  6,  assign- 
ing 6  to  the  point  (1,  0).  The  root- 
points  are  at  the  positions  1,  3,  5.  The 
points  numbered  1,  5  have  a  common 
.abscissa  which  is  obviously  ^ ;  and 
their  ordinates  are 

Hence  the  three  roots  are 

Ex.  2.  Solve  the  equation  z^^  =  1. 

If  an  inscribed  regular  decagon  and  an  inscribed  regular  hexa- 
gon have  a  common  vertex  at  A  (Fig.  8)  and  their  adjacent  ver- 
tices be  at  P  and  Q  respectively,  PQ  is 
the  side  of  a  regular  pentadecagon.  Take 
AB  =  PQ ;  then  A  and  B  are  two  adja- 
cent vertices  of  a  regular  pentadecagon. 
Let  the  co-ordinates  of  P,  §,  and  B  be  x, 
y,  u,  V,  and  x',  y'. 

Then         m,  v  =  i,  JV3» 

x,y  =  i(l  +  V5),  i V(10  -  2 V5), 
and  x',  y'=  ux  -\-  vy,  vx  —  uy.  [Art.  424.] 
Therefore  the  co-ordinates  of  the  root-point  of  i^^^  =  1  which 
is  adjacent  to  A  are 


Fig.  8. 


iC' 


=  H1+V5+V(30-6V5)}, 


2/'  =  HV3+Vl5-V(10-2V5)}, 
and  the  corresponding  root  of  z^^  =  1  is 
(0  =  x'  +  iy'. 
The  fifteen  roots  of  z^^  =  1  are  therefore  w,  w^,  w*,  ♦••,  w^^ 


GRAPHICAL  METHODS.  623 

Binomial  equations  of  the  form  2"  =  ±  1  are  a  special 
type  of  reciprocal  equations.     [See  Art.  449.] 

EXAMPLES  CVI.  V- 

Solve  the  following  equations  ; 

1.  z^  =  27.                  6.   z^  =  --l.  9.  2^8  =-16. 

2.  ^4  =  4.                     6.    ^  =  5.  10.  2:i'>  =  1024. 

3.  z^  =  -l.                 7.   06  =  -27.  IL  2io  =  -16. 

4.  05  =  32.                    8.    08  =  625.  12.  0I6  =  1. 

Verify  the  following  statements  : 

13.  02n+i  =  1  has  the  real  root  +  1,  but  no  others. 

14.  02n+i  =  _  1  has  the  real  root  —  1,  but  no  others. 

16.   02n  =  1  has  the  two  real  roots  +  1  and  -  1,  but  no  others. 

16.  All  the  roots  of  z-**  =  -1  are  complex. 

17.  If  m  be  prime  to  n,  the  equations  0"*  =  1  and  0"  =  1  have 
no  common  root  except  unity. 

18.  If  n  be  a  prime  number  and  a  any  one  of  the  complex  roots 
of  0"  =  1,  then  a,  a%  o^,  •••,  a^^  are  the  n  roots  of  0**  =  1. 


624 


DERIVATIVES. 


CHAPTER  XLIV. 


DERIVATIVES. 


428.   The  Difference  Katio.     Let  f(x)  denote  the  rational 
integral  function 

The  graph  of  this  function  is  a  curve,  ^nd 

is  its  equation;  values  of  x  are  abscissae,  values  of  y 
ordinates.  For  a  second  pair  of  co-ordinates  x^,  y^  there 
exists  the  similar  equation 

2/i=/(^i)- 
Let  a?!— ic  and  yi—yhe  denoted 
by    Ax   and    Ay    respectively.* 


Pi 

1 

y 

Ay 

AX 

M 
Fig.  9. 


Then 


and 


Ay=f(x  +  Ax)-f(x) 

Ay      f(x-\-A*x)~f(x) 


Ax  Ax 

This  fraction  is  called  the  difference-ratio  of  the  function 
f(x).     In  the  figure  (Fig.  9)  with  OX,  0  Y  as  axes  and 


*  A  and  the  letter  following  it  are  regarded  as  inseparable. 
Thus  Ax  X  Ax  =  (Ax) 2,  and  this  cannot  be  written  A^x^,  Similarly 
Ax  X  Ay  cannot  be  written  A'^xy. 


DERIVATIVES.  625 

PSPi  a  segment  of  the  curve,  the  co-ordinates  of  P  and 
Pi  are 

OM,  MP,  0M„  M,P,  =  X,  y,  x„  y^ 

and  ^  =  ^. 

Ax      PR 

In  the  triangle  PRPi,  right  angled  at  R,  the  ratio 
RPy/PR,  that  is,  the  ratio  of  the  opposite  side  to  the 
adjacent,  is  called  the  tangent  of  the  angle  RPPi,  and  is 
designated  by  the  abbreviation  tan  RPP^ ;  hence 

^  =  tan  i2PPi. 

Ax 

Thus,  the  difference-ratio,  corresponding  to  any  two 
points  on  the  curve,  is  equal  to  the  tangent  of  the  angle 
which  the  secant  line  through  the  points  makes  with  the 
aj-axis. 

429.  The  Derivative.  Let  the  values  Xi,  y^  now  approach 
the  values  x,  y ;  then  Ax  and  Ay  both  approach  zero,  and 

the  difference  ratio  tends  to  assume  the  form  -,  whose 

limit,  however,  we  may  propose  to  determine.  In  the 
figure.  Pi  approaches  P,  the  secant  line  turns  about  P 
and  approaches  coincidence  with  the  tangent  line  at  P. 
Denoting  by  <^  the  angle  which  this  tangent  line  makes 
with  the  aj-axis,  we  have,  in  the  limit,  when  Aa*  =  0  and 
Ay  =  0, 

limit    A^  ^  limit  f(x-^Ax)-f(x)  ^  ^^^ 

This  limit  is  a  new  function  of  x.     It  is  called,  the 
derivative  of  f{x)  with  respect  to  the  variable  x,  and  is 
2s 


626  DERIVATIVES. 

denoted  by  f(x).  It  is  equal  to  the  tangent  of  the  angle 
which  the  tangent  line  at  the  point  {x,  y)  makes  with  the 
X-axis.     To  differentiate  a  function  is  to  find  its  derivative. 

Ex.  1.    Find  the  derivative  of  oj^. 
The  difference-ratio  is 

fix  +  Ax)  -/(x)  ^  (X  -^  Ax)2  -x^^.^^      ^.  Ag, 
Ax  Ax  ^Ax 

Discarding  the  unity  factor,  and  passing  to  the  limit,  we  have 
/'(x)=2x. 

In  the  differential  calculus  the  derivative  is  a  function 
of  far-reaching  importance,  and  in  that  subject  it  calls 
for  extended  discussion  and  the  setting  up  of  a  large 
number  of  general  and  special  formulae.  Here  we  shall 
only  need  to  consider  a  few  special  cases.  In  particular, 
we  need  to  be  able  to  write  down  the  derivative  of  a 
rational  integral  function. 

430.  Derivative  of  a  Constant.  Since  a  constant  does  not 
undergo  any  change  in  consequence  of  the  changes  of 
other  quantities,  its  difference-ratio  is  always  zero,  and 
hence  its  derivative  is  also  zero. 

431.  Derivative  of  a  Sum.  Let  /^(x)  and  f2{x)  be  two 
different  functions,  and  let 

f(x)=Mx)+Mx). 

The  difference-ratio  of  this  sum  is 

f(x  -\-  Aa?)  -f{x)  ^Mx  +  Ax)  -f /2 (a;  +  Ao;)  -/^ (x)  -f^ (a?) 
Ao;  Aa; 

_/i(a;  -f  Ax)  -fi(x)      f^jx  +  Aa;)  -f^jx) 
Ax  Ax 


DERIVATIVES.  627 

Hence,  passing  to  the  limit  for  Ax  =  0, 

That  is,  the  derivative  of  the  sum  of  any  two  functions 
is  equal  to  the  sum  of  their  derivatives. 

In  this  proof  it  has  been  taken  for  granted  that  the 
limit  of  the  sum  of  any  two  quantities  is  equal  to  the 
sum  of  their  limits. 

432.  Derivative  of  a  Product.  Let  f  (x)  and  /g  (x)  be  any 
two  functions,  and  let 

f(x)=Mx).Mx). 

The  difference-ratio  of  this  product  is 

f(x  +  Ax)  -fix)  ^  /i  (a;  +  Aa;)  ./^(a;  +  Ax)  -f^(x)  -f.jx) 
Ax  Ax 

^  [f(x  +  Ax)  -f(x)^f2(x  +  Ax) 
Ax 
lA{x  +  Ax)-f,(x)^f{x) 
"^  Ax 

Hence,  passing  to  the  limit, 

/'  (x)  =Aix)A'(x)  +Mx)f,'ix). 

That  is,  the  derivative  of  the  product  of  any  two  quan- 
tities is  equal  to  the  second  into  the  derivative  of  the  first 
plus  the- first  into  -the  derivative  of  the  second. 

Here  again  it  has  been  taken  for  granted  that  the  limit 
of  the  sum  of  two  quantities  is  equal  to  the  sum  of  their 
limits. 


628  DERIVATIVES. 

If  a  be  a  constant,  we  have,  as  a  corollary  of  the  last 
formula,  that  if 

f{x)  =  af^{x), 

then  f(x)=af^{x). 

Hence,  the  derivative  of  the  product  of  a  constant  and  a 
function  is  equal  to  the  constant,  multiplied  by  the  deriva- 
tive of  the  function. 

433.  Derivative  of  a  Power.  Let  f(x)  =  x\  The  differ- 
ence-ratio is 

f(x  +  Ao;)  —f(x)  _(x-\-  Axy  —  x"" 
Ax  ~  Ax 

But  (x  +  ^xy  =  a;"  +  nx^-^Ax  +  MAx^j 

where  JHf  is  a  multinomial  in  x  and  Ax.     Hence 

f(x-\-Ax)—f(x)       .     ^  ,  .  , -.^   .  Aaj 
^  ^  "^ — ^^-^^  =  (nx^'-^+MAx)  -— . 

Ax  ^  ^  Ax 

This  last  expression  differs  from  na;"~^  by  an  arbi- 
trarily small  quantity  when  Ax  is  made  sufficiently  small. 
Therefore,  in  the  limit,  when  Aic  =  0, 

f'(x)  =  nx'^~\ 

This  proof  presupposes  that  n  is  a  positive  integer, 
though  the  formula  is  true  for  all  values  of  n. 

Thus,  the  derivative  of  the  n^^  power  of  the  independent 
variable  (n  =  a  positive  integer)  is  n  times  the  (n  —  1)*^ 
power  of  the  variable. 

This  rule  is  also  applicable  to  finding  the  derivative  of 
(x  ±  ay.     [See  Ex.  2  of  Art.  435.] 


DERIVATIVES.  629 

434.  Derivative  of  f(x-\-a).  If  a  be  a  constant,  the 
difference  between  two  values  of  x-\-  a  is  equal  to  the 
difference  between  the  corresponding  values  of  x,  for  a 
does  not  change.     Thus 

A  (a;  -h  a)  =  Ax, 

and  if /(a;  +  a)  be  any  function  of  a;  -f  a,  then 

Af{x  +  a)  =/[a;  +  a  4-  A(a;  +  a)]-/(a;  +  a) 

=zf(x-{-a  +  Ax)  -fix  +  a) ; 

that  is,  the  difference-ratio,  and  therefore  also  the  deriva- 
tive, of  f{x  -h  a)  is  the  same  whether  a;  +  a,  or  a;  alone, 
be  regarded  as  the  independent  variable.  Hence  the 
following  theorem : 

The  derivative  of  f{x  +  a)  with  respect  to  x  =  the  deriva- 
tive of  fix -\-  a)  with  respect  to  x+a=f\x  +  a).  [See 
Ex.  2,  Art.  435.] 

435.  The  foregoing  formulae  include  all  the  cases  of 
differentiation  that  will  present  themselves  in  the  follow- 
ing pages,  but  a  few  examples  are  added  in  further  illus- 
tration of  the  subject. 

Ex.  1.   Find  the  derivative  of  the  quotient  of  any  two  functions. 

Then  fix)'f2{x)  =  fiix), 

and,  by  differentiation  [Art.  432], 

fix)  -Mx)  -^f^'ix)  .fix)  =/l'(«), 


6S0  DERIVATIVES. 


But  /(x)  = 


/2(^) 


Ex,  2.    Find  the  derivative  of  {x  —  a)^. 
The  difference-ratio  is 

{x  +  Ax  -  ay- {x  -  ay _  n{x-  ay-^^x  +  M^y?' 
Ax  ~  Ax  ' 

where  iW  is  a  multinomial  in  (x  —  a)  and  Ax.  In  passing  to  the 
limit,  with  Ax  =  0  everything  disappears  except  n{x  —  a)"-^,  which 
is  the  required  derivative. 

Ex.  3.    Find  the  derivative  of  a". 
The  difference-ratio  is 

=  a* 


Ax  Ax 

In  Art.  307,  Ex.  1,  it  was  shown  that 

limit  «^i:zl  =  lna. 

AX:^0         Ax 

Hence,  derivative  of  a==  =  a*  In  a. 
Hence  also,  derivative  of  e'  =  e*. 

EXAMPLES   CVII. 
Find  the  derivative  of  each  of  the  following  functions : 

1.  x2  +  x3.  8.   x"'(x'*  +  a).  jg    J^. 

2.  3x2  +  2x3.  9.    (x+l)3.  ^^ 


3.    (x+l)(x2-2).  10. 


X  +  1  14. 


x'^ 


4.    5x2(3x+l).  ^  15.  I 

6.      X2  +  2  &X  +  C.  -1  17  b\Qg  rf 

12.  i  +  a;.  ^'      ^ 

7.   ax«  4-6.  ic  18.  ln(x  +  a). 


DERIVATIVES.  631 

436.  If  fix)  have  the  factor  (x  +  a)",  the  derivative  of 
fix)  has  the  factor  (x  +  a)"~^. 

For,  if  /(a;)  =  (ic  +  a)"<^(a;), 

then    f'(x)  =  n(x-\-a)''-^<ti{x)  +  (x  +  a)''<f>'(x)      ' 
=ttB  +  a)"-i  \n<f,  (x)  +  (a;  +  a)  cf>\x)l, 

437.  Equal  Boots.  The  following  is  an  immediate  cor- 
ollary of  the  foregoing  theorem : 

If  f(x)  =  0  have  k  roots  equal  to  a,  /' {x)  —  0  has  k—1 
roots  equal  to  a. 

It  is  evident,  from  the  form  of  /'  (a;),  that  x -\-  a  will 
not  appear  as  a  factor  in  /'  {x)  unless  it  appears  at  least 
twice  as  a  factor  in  f{x).  Hence  the  necessary  and  suffi- 
cient condition  that  fix)  =  0  shall  have  equal  roots  is 
that  fix)  and  /'  (x)  shall  have  a  common  factor  involv- 
ing x. 

To  find  the  equal  roots,  obtain  the  highest  common 
factor  of  fix)  and  /'  (a;)  and  solve  the  equation  obtained 
by  putting  this  factor  equal  to  zero. 

Ex.  The  H.C.F.  of  jc*  -  Qx^  +  4  a; -I- 12  and  its  derivative 
4  X*  —  18  X  4-  4  is  found  to  be  x  —  2.  Hence  (x  -  2)*  is  a  factor 
of  X*  —  9x2 -f- 4x -f- 12,  the  other  two  factors  of  which  are  then 
obtained  by  division  and  inspection.     They  are  x  +  1  and  x  +  3. 

438.  Maxima  and  Minima.  If  fix)  increase,  attain  a 
value  /(a)  and  then  decrease,  /(a)  is  said  to  be  a  maxi- 
mum  value  of  fix).     If  fix)  decrease,  attain  a  value 


632 


DERIVATIVES. 


/(a?) 


Fig.  10. 


/(/8)  and  then  increase,  f(/3)  is  said  to  be  a  minimum 

value  of  f(x). 

Suppose  f(x)  to  be  a  rational  integral  function  of  x, 

and  let  its  graph  be  constructed.  It  may  have  maximum 
and  minimum  ordinates,  as  at  A 
and  B  (Fig.  10). 

As  the  point  (x,  y)  passes  along 
the  curve  from  left  to  right,  x 
increases  and  Ax  is  positive. 
During  this  movement  f(x)  in- 
creases or  decreases  according 
as   A?/  is   positive   or   negative, 

that   is,   according   as   the   difference-ratio   Ay/Ax,   and 

therefore  the  derivative  f'(x),  is  positive  or  negative. 

In  fact,  briefly, 

f(x)  is  an  increasing  or  a  decreasing  function  according 
as  f  (x)  is  positive  or  negative. 

If,  then,  f  (x)  pass  from  positive  to  negative  values, 
becoming  zero  at  the  instant  of  change,  f{x)  will  first 
increase  and  then  decrease,  will  attain  in  fact  a  maximum, 
at  the  instant  when  /  {x)  changes  sign.  Similarly,  f{x) 
will  attain  a  minimum  when  /'  (x)  passes  from  negative 
to  positive  values.     Hence 

If  f  (a)  =  0,  and  if  in  passing  through  this  value  f  (a) 
pass  from  positive  to  negative  values,  then  /(a)  is  a 
maximum. 

■  Jf  f  (o.)  =  ^?  ^1^^  if  ^^  passing  through  this  value  /'(a) 
pass  from  negative  to  positive  values,  then  /(a)  is  a 
minimum. 


DERIVATIVES. 


633 


The  theory  of  maxima  and  minima  is  an  important 
aid  in  tracing  graphs  of  functions. 


Ex.  1.  Construct  the  graph  of  aj^  —  3  x,  determining  the  maxi- 
mum and  minimum  ordinates. 

The  derivative  of  x^  -  3  a;  is  3  ic2  -  3,  and  the  roots  of  3  x2_3  =:  o 
are  1  and  —  1.  The  maximum  and  minimum  ordinates,  if  there 
be  any,  must  therefore  be 

13_3.i=_2  and  (_  l)3_  3(_  i)  =  2. 


But  it  is  easy  to  see  that  x^  —  1, 
the  variable  part  of  the  derivative, 
changes  from  a  negative  to  a  posi- 
tive quantity  when  x  passes  through 
1 ,  and  from  a  positive  to  a  negative 
quantity  when  x  passes  through  —1 ; 
hence  —2  is  a  minimum  ordinate  and 
2  is  a  maximum  ordinate. 

The  graph  of  the  function  is  rep- 
resented in  Fig.  11. 


{-2,-2) 


Ex.  2.   Find  the  maximum  and  minimum  values  of  x^  —  9  x. 

The  derivative  of  x'^  —  9  x  is  3  (x^  —  3)  and  the  values  of  x 
which  make  it  vanish  are  —  V3  and  +  \/3.  The  corresponding 
maximum  and  minimum  ordinates  are  6\/3  and  —6  VS.  The 
graph  is  that  of  Fig.  4,  Art.  421. 


439.  Successive  Derivatives.  The  successive  results 
obtained  by  repeating  the  process  of  differentiation  are 
called  successive  derivatives ;  if  the  process  be  once,  twice, 

•  ••  (n  —  1)  times  repeated,  the  result  is  the  second,  third, 

•  •  •  ii**"  derivative.  These  successive  derivatives  of  an 
J{x)  are  denoted  by  f'(x),  f'(x),  f'(x)y  •••/'*X^)?  respec- 
tively.    For  example,  if 


634 .  DERIVATIVES. 

f{x)  =x% 

then  f(x)  =  nx^-^j 

f"(x)  =  n{n  -  l)x''-^, 

f"'(x)  =  n(n  -  l)(n  -  2)af-«, 


f(-\x)  =  n{n  -  l)(n  -  2) ... 3  •  2  •  1, 

and  all  the  higher  derivatives  are  zero.  f(x)j  f'(x)j 
••'P''\x)j  are  also  called  the  first,  second,  -^  n^  derived 
functions  of  f(x). 

EXAMPLES   CVIII. 
Find  the  third  derivative  of  each  of  the  following  functions : 
1.    _^!_.  2.   -^L-.  3.        ^' 


x^  -  a2  x^-  a2  ic2  _  ^2 

Find  the  n}^  derivative  of  each  of  the  following  functions  : 
4.   -J_.  6.   i+^.  6.    (1+^. 

1  —X  I  —X  1  —X 

Solve  the  following  equations,  each  of  which  has  equal  roots  : 

7.  x3  +  15  x2  -  33  X  +  17  =  0. 

8.  «*  -  6  x3  +  11  ic2  -  4  a;  -  4  =  0. 

9.  x5  -  3  x*  -  jc3  +  7  ic2  -  4  =  0. 

10.  x6_i2a:2-16  =  0. 

11.  2  x*  -  9  x3  +  6  x2  +  20  X  -  24  =  0. 

12.  x5  -  3  x4  +  5  x3  -  7  x2  +  6  X  -  2  =  0. 

Note.  — In  examples  11  and  12  find  the  highest  common  factor 
of  the  first  and  the  second  derivatives. 


DERIVATIVES.  635 

Construct  the  graphs  of  the  following  equations  : 

13.  y  =  x^-3x^-9x-5.  16.    y  =  x^- 2x^^-5. 

14.  xy  -x"^-  y-l=0.  17.   y  =  2  yfi  -  6  xf^ -\- 5  x'^ -\- 1. 

15.  x2  +  ?/2  -  2  cc  =  0.  IS.   xy  -  y  +  1  =  0. 

19.  It  is  required  to  construct,  with  the  least  possible  amount 
of  sheet  metal  (thickness  of  metal  given),  a  cylindrical  vessel, 
open  at  the  top,  which  shall  contain  a  given  quantity  of  liquid. 
Determine  (in  terms  of  its  volume)  the  exact  (interior)  dimensions 
of  the  vessel. 


636  Taylor's  theorem. 


CHAPTER  XLV. 

Taylor's  Theorem. 
440.   Let/(aj)  denote  the  rational  integral  function 

Then  f(x  -j-  h)  is  a  rational  integral  function  of  h  and 
can  be  arranged  in  ascending  powers  of  h,  thus : 

f(x  +  h)  =  A  +  A^  +  A2h^  H +  AJi% 

in  which  A,  ^i,  etc.,  involve  x,  but  not  h.  Differentiate 
each  member  of  this  equation  with  respect  to  h,  regarding 
X  as  constant,  and  repeat  the  process  on  each  successive 
equation  until  /^"^^  +  ^0  ^^  obtained.     The  results  are 

f(x  +h)    =  A  +  2  A,h  +  3  A,h'  +  -, 

f'(x  -^h)   =  2  A  +  2  .  3  ^/i  4-  3  •  4  AJv"  +  ••-, 

f"(x  +  h)  =2.3^  +  2.3.4^/1  +  3.4.5^/124-..., 

/(")(a;  +  /i)  =  1.2.3...nA. 

In  these  equations  put  7i  =  0 ;  then 

/'(^)      =2  A,  f"(x)=l-2.3A„ 

/(-i)(a!)  =  (n-l)!A-i,  /">(«)  =  »!  A- 


637 

Thus  the  values  of  Aq,  A^,  A2,  •  •  •  A^  are  known,  and 

f(x  +  h)  =f{x)  +  hf(x)+^f'(x)  +  •..  +2/"X«^)- 

This  formula  is  known  as  Taylor's  Theorem.  It  here 
appears  in  a  special  form,  applied  to  a  rational  iiitegral 
function.  The  values  of  f'(x),  f'(x),  etc.,  are  now  found 
by  differentiating  successively  the  original  multinomial, 

f(x)     =  aox""  +  ch^f-^  H h  a^-^x  +  a„, 

and  its  derivatives.     Thus, 

f(x)    =  noox''-^  +  (n  —  l)(Xia;"-2  H h     a*-^ 

f\x)    =n(ri-l)aoa;"-2+...  -f2a''-2, 

/(-)(a;)  =  1.2.3...7iao. 

The  student  may  verify  these  results  by  writing  x-\-h 

for  x  in.aoo;**  +  ajOj"-^  + \-  a^,  developing  the  binomials 

(x  +  hy,  (x  4-  hy~'^,  etc.,  and  collecting  the  coefficients 
of  the  various  powers  of  h. 

Ex.  1.  Substitute  1  +  h  for  x  in  o^  —  Sx^  +  a;  +  2  and  develop 
the  result  in  terms  of  powers  of  h. 

Let  fix)  =  x8  -  3x2  +  X  +  2. 

Then     /'(x)=  3x2  -  6x  +  1,   /"(x)=6x-6,    /'"(x)  =  6, 

and  all  the  higher  derivatives  are  zero. 

The  substitution  of  1  in  these  functions  gives 

/(1)=1,  /'(l)  =  -2,  /"(!)=  0,  /•"(!)=  6. 

.'.f{l  +  h)=l-2h  +  h^ 

=  (l-h){l-h-h^). 


638  Taylor's  theorem. 

Ex.  2.    Develop  (1  +  x)^  in  ascending  powers  of  x. 

Let  /(x)  =  x^. 

Then  /'(x)  =  5x*,  /"(x)  =  20x3,  /"'(x)=60x2, 

/iv(x)  =  120  X,  and  /^(x)  =  120. 

.•./(l  +  x)  =  /(l)  +  x/'(l)+ixV"(l)+- 

=  1  +  5X  +  -2f  X2  +  -6^X3  +  V^X*  +  X^ 

=  1  +  5  X  +  10  x2  -f  10  x3  +  5  x4  +  x5. 

441.  Oontinuity.  It  has  been  seen  [Art.  284]  that 
variables  and  functions  may  change  continuously,  or 
discontinuously.  The  continuity  of  a  function  /(a?),  for 
a  given  value  of  x,  is  tested  by  observing  the  behavior 
of  the  difference 

f{x^K)-f{x\ 

where  fi  is  an  arbitrarily  small  quantity  and  either  pos- 
itive or  negative.  If  this  difference  remain  permanently 
smaller  than  an  arbitrarily  small  quantity  8,  while  h 
passes,  through  either  positive  or  negative  values,  into 
the  value  zero,  the  function  fix)  is  continuous  for  the 
value  X.     Thus  y?  is  always  a  continuous  function ;  for 

{x-\-}if-y?=^2xh^h?, 

and  this  is  zero  when  /i  =  0  whether  li  pass  into  zero 
through  positive  or  negative  values.  But  l/(a;  —  1)  is  a 
discontinuous  function  at  a;  =  1 ;  for 

1 1     ^  -Ti 

x-\-}i  —  \      x  —  1      {x  —  V){x^'h  —  V) 

and  if  a;  =  1  this  is  —  /i/(0  •  /i),  and  is  —  oo  when  h  ap- 
proaches zero  through  positive  values,  but  becomes  4-  oo 
when  /i  passes  into  zero  through  negative  values. 


Taylor's  theorem.  639 

442.  Oontinuity  of  /(«).  Suppose  the  variable  x  to 
undergo  continuous  change,  and  let  a  and  a-\-h  be  any 
two  finite  values  of  x.  If  f{x)  be  a  rational  integral 
function  of  a;,  then  by  Taylor's  theorem, 

/(a  +  h)  -/(a)  =  A/'(a)  +|^/"(a)  +  •.•  +^/<"'(<')- 

If  h  be  arbitrarily  small,  the  difference /(a  +  h)  —f{a) 
is  evidently  itself  arbitrarily  small.  Hence,  by  the  cri- 
terion of  Art.  441,  the  function  f(x)  changes  continu- 
ously when  X  undergoes  continuous  change. 

In  this  proof  a  and  a-\-h  have  any  finite  or  zero  values. 
Hence  the  rational  integral  function  f(x)  is  continuous  for 
all  finite  and  zero  values  of  the  variable. 

This  theorem  shows  us  that  the  graph  of  a  rational 
integral  function  is  a  continuous  line. 

Corollary.  Tlie  derivatives  of  a  rational  integral  func- 
tion (being  themselves  rational  and  integral)  are  continuous 
for  all  finite  and  zero  values  of  the  variable. 

443.  Rolle's  Theorem.  Iff(x)  assume  the  same  value  for 
two  different  values,  a  and  b,  of  x,  the  derivative  f  (x)  will 
vanish  for  some  value  of  x  between  a  and  b. 

As  X  passes  continuously  from  a  to  6,  f(x)  passes  con- 
tinuously from  f{a)  to  f{b)  [Art.  442].  But,  by  hypothe- 
sis, /(a)  =f{b),  and  since  f(x)  varies,  it  must  attain,  in 
the  interval  a,  b,  at  least  once,  either  a  maximum  or  a 
minimum  value  [Art.  438].  Therefore /'(a;)  must  vanish, 
at  least  once,  for  a  value  of  x  between  a  and  b. 

Hence  also,  Iff{x)  —  0  have  two  distinct  real  roots,  a  and 
b,  f  (x)  =  0  will  have  at  least  one  real  root  between  a  and  b. 
This  is  EoUe's  Theorem. 


640  Taylor's  theorem. 


EXAMPLES  CIX. 

Show  that  the  following  functions  are  discontinuous  for  the 
value  x  =  1 : 

1.   -^ —  2. — -.  3.    In  (X  -  1). 

x^-1  (X  -  1)2  ^  ^ 

Show  that  the  following  functions  are  continuous  for  all  finite 
values  of  x : 

4.    (x-a)".  5.    —^ —  6.   a^. 

^  ^  x2  +  1 

Develop  the  following  functions  into  infinite  series  in  ascending 
powers  of  x,  by  Taylor's  theorem,  and  assign  the  range  of  values 
of  X  for  which  the  series  are  convergent : 


Y    x  +  1 

■    X-  1 

9.           1        . 

V(^  +  1) 

11           1 

(X  -  1)2 

8.    e-+\ 

10.    a^-\ 

12.    In  (x.+  1), 

13.  Show,  by  the  aid  of  Rolle's  theorem,  that  for  the  equation 
/(x)  =  0,  of  the  third  degree,  to  have  all  its  roots  real  and  unequal, 
it  is  necessary  and  sufficient  that  the  roots  a,  j3,  of  f'(x)  =  0,  shall 
be  real,  and  that 

/('^)-/(^)<0. 

14.  Apply  the  theorem  of  example  13  to  x^  +  gx  +  »'  =  0,  and 
show  that  the  condition  there  stated  is  fulfilled  if 

4  ^3  _i_  27  7-2  <  0. 


TRANSFORMATION.  641 


CHAPTER  XLVI. 

Transformation. 

444.  When  a  multinomial  in  x  is  so  changed  as  to  be- 
come a  multinomial  in  some  function  of  x,  or  of  another 
variable,  it  is  said  to  be  transformed.  The  general  type  of 
all  the  important  transformations  in  the  elementary  theory 
of  equations  is  the  so-called   homographic  transformation, 

whereby  x  is  replaced  by  a  function  of  the  form  ^^      \^- 

yZ  -^6 

This  general  type  of  transformation  is  a  combination 
of  the  three  special  types 


z  H-  hj  kz,  and  - ; 
z 


for,  identically 


yZ-^8       y  y^  Z -\- S/y      y      Z-\-h' 

where  k  =  PI  ~^    and  h  =  S/y ;  and  the  successive  sub- 
stitutions       ^ 

X^  ^^^  Z  ~\~  flm    X2  ^—  —  J    X^  —  f€X2)    X^  —  fl/g  ~|~  — 

^  y 

lead  directly  to 

a  ,      k 

y       Z  +  h 

The  effect  of  the  three  special  types  of  transformation  on 
a  function  is  to  change  it  to  a  function  of  a;  -f-  h,  or  of  kx, 
2x 


642  TRANSFOEMATION. 

or  of  -;  and  on  an  equation  the  effect  is :  (1)  to  increase 

its  roots  by  h,  or  decrease  them  by  —h\ih  be  negative ; 
(2)  to  multiply  its  roots  by  k,  or  divide  them  by  1/k  if 
A;  <  1 ;  or,  (3)  to  change  the  roots  into  their  reciprocals. 
The  three  types  will  be  taken  up  separately. 

445.    To  transform  f(x)  into  a  function  ofx  —  a. 
In  Taylor's  theorem  applied  to  the  function  /(a  +  h) 
[Art.  440],  let  ic  —  a  take  the  place  of  h ;  then 

f{a  +  X-  a)  =f{x)  =/(a)  +  (x  -  a)/(a) 

+  ^>(a)+-+^^/»'(a), 

which  is  f(x)  expressed  in  terms  of  the  powers  of  x  —  a. 
It  is  the  transformed  function  required;  the  new  coeffi- 
cients are  functions  of  a. 

In  Art.  416  it  was  seen  how  these  coefficients  may  be 
computed  by  simple  and  successive  division  by  x  —  a, 
and  the  rule  for  the  computation  is  there  given. 

The  proof  of  the  rule  was  given  for  a  multinomial  of 
the  fourth  degree.  The  student  will  now  find  no  difficulty 
in  rehabilitating  the  proof  for  the  general  case. 

Ex.  1.  Transform  x*  +  11  x^  +  41  a:2  +  61  x  +  30  into  a  function 
of  X  +  2. 

The  new  coefificients  are  1,  3,  —  1,  —  3,  0,  and  the  transformed 
function  is 

{x  +  2)4  +  3  (x  +  2)3  -  (x  +  2y-S(x  +  2). 

This  transformation  was  constantly  used  in  the  applica- 
tions of  Horner's  method,  and  the  student  is  referred  to 
Art.  416  for  other  examples. 

It  was  also  used  in  Art.  408  for  getting  rid  of  the 


TRANSFORMATION.  643 

second  term  of  the  general  cubic  oc^  -\-  px^  +  qx -\- r,  and 
it  may  be  used  with  similar  effect  on  any  multinomial. 

Ex.  2.  Transform  ax*  +  bx^  +  cx'^  +  dx  +  e  into  a  function  of 
X  —  a  and  assign  a  value  of  a  that  will  make  the  coefficient  of 
(x  —  a)3  vanish. 

Here  /(a)     =  aa"^  +  ba^  +  ca^  +  da  +  e, 

/'(a)    =4aa3  +  36a2  +  2ca  +  (i, 

/"(a)  =12aa2  +  6&a  +  2c, 

/'"(a)  =  24aa  +  6  6, 

/iv(a)  =  24a. 
Hence  the  transformed  function  is 

a  (X  -  a)*+(4  aa  +  6)(x  -  0)84-  (6  aa^  4-  3  6a  +  c)(x  -  a)2 

+  (4  aa8  +  3  6a2  +  2  ca  +  d)  (X  -  a) 

+  aa*  +  6a3  -f  ca"^  ^  da  +  e. 

The  coefficient  of  (x  —  o)^  will  vanish  if  a  =—  b/(4a).  The 
third  term  will  drop  out  if  a  satisfy  the  equation 

6  aa2  +  3  6a  +  c  =  0, 
the  fourth  term  if 

4  aa8  +  3  6a2  +  2  ca  +  (^  =  0. 

When  the  multinomial  of  the  n^  degree, 

f(x)  =  aox""  +  ttio;"-^  H h  a^_ix  +  ««, 

is  transformed  in  this  fashion,  the  coefficient  of  (a;— a)"~^ 
in  the  transformed  function  is 

T-^^TTT  /"-'^  (a)  =  oi  +  na^ 
{n  -  1) ! 

the  coefficient  of  (x  —  a)"~^  is 

1 f ("-2)  (a\  =  a„  4-  (n  -  V\  a,a  4-  ^(^~^) 


/("-2)  (a)  =  a,  +  (n  -  1)  a^a  +  '1111^^  a^\ 


(71-2)!"         '"        ^  ^  ''    ^     '         2 


644  TRANSFORMATION. 

and  so  on.  Any  term  in  the  transformed  function  may 
therefore  be  made  to  disappear  by  solving  an  equation  in 
a,  of  the  first  degree  if  it  be  the  second  term  that  disap- 
pears, of  the  second  degree  if  it  be  the  third  term,  and 
so  on. 

446.  To  transform  f(x)  into  a  function  of  kx. 

To  achieve  this  transformation  we  have  only  to  replace 
ic  by  -  •  kx.     If  the  given  function  be 

a(fi(f  +  ajps''-'^  H h  a„_ia;  +  a^ 

the  transformed  function  is 

^ (kxy  +  y^,  (kxy-' -\-'"-\-^'Qcx)  +  a„. 

By  means   of  this   transformation   troublesome   frac- 
tional coefficients  may  be  got  rid  of. 

Ex.    1.    Transform   x^  -\-  ^x"^  —  ^x -{-  ^^  =  0   into   an   equation 
whose  coefficients  are  all  integral. 

Replace  x  by  kx  and  multiply  by  k^.    The  equation  becomes 

where  y=:kx;  and  if  A;  =  6, 

this  is  y3  _}.  10  2/2  _  20  2/  +  9  =  0. 

447.  Note  the  special  case  k  =  —  l.     We  here  substi- 
tute —  (—  0?)  for  X  and  obtain 

(-  1)X(-  «')"+(-  l)"-'«i(-  ^y-'  + «n-i(-  ^)  +  a„, 

or  ±  ao2/"  =F  ai^/""^  ± a„-i2/  +  «„, 


TRANSFORMATION.  645 

wherein  y  —  —  x.  Observe  that  the  signs  of  the  terms 
become  alternately  plus  and  minus.  Applied  to  an  equa- 
tion, this  transformation  produces  a  new  equation  whose 
roots  are  those  of  the  old  with  their  signs  changed. 
It  is  useful  in  the  computation  of  negative  roots  by 
Horner's  method;  the  transformation  takes  place,  of 
course,  before  the  computation  begins. 

A  simple  rule  for  writing  the  transformed  equation  in 
this  case  is  obvious.     It  is : 

Change  the  sign  before  every  other  terrriy  beginning  with 
the  seco7id. 

448.  Eeciprocal  Equations.  It  is  evident  that  a  function 
of  X  cannot,  by  any  transformation,  be  written  as  a  func- 
tion of  1/x,  and  that  therefore  the  third  type  of  trans- 
formation [Art.  444]  necessarily  introduces  a  new  variable 
Zy  connected  with  the  old  by  the  relation  z  =  l/x. 

Applied  to  an  equation,  say  the  general  equation 

aoa?"  -f  aix""-^  + h  a„_ix  -f  a„  =  0, 

this  transformation,  x  =  1/z,  after  multiplication  by  2", 
leads  to  a  reciprocal  equation 

aX  +  ««-i2;""^  -\ 1-  ^iZ  +  a©  =  0, 

and  the  two  equations  obviously  stand  to  one  another  in 
the  following  relation  of  reciprocity :  The  coeflBcient  of 
af  in  either  is  the  same  as  the  coefficient  of  a?""'"  in  the 
other,  and  the  roots  of  the  one  are  the  reciprocals  of  the 
roots  of  the  other. 

449.  An  equation  is  its  own  reciprocal,  that  is,  self- 
reciprocal,  if  it  remains  unchanged  when  x  is  replaced 


646  TRANSFORMATION. 

by  l/x,  after  its  terms  have  been  multiplied  by  the 
proper  power  of  x.  Thus,  a^  +  2a^  —  2a7  — 1  =  0  is  a 
self-reciprocal  equation,  or  simply  a  reciprocal  equation."* 

If  the  general  equation  of  the  n^^  degree  satisfy  this 
test,  the  two  equations  of  the  last  article  must  be  identi- 
cal, the  necessary  and  sufficient  condition  for  which  is 
[Art.  328] 

^  =  _^  =  _^  =  ...  =  ^!^  =  ^. 

This  condition  may  be  expressed  more  briefly  thus, 
^  =  ^'-  =  e,  [r  =  0,l,2,...7i] 

where  e  is  either  +1  or  —  1,  but  is  the  same  for  all  the 
ratios. 

This  formula  furnishes  the  complete  criterion  for  a  recip- 
rocal equation. 

Ex.  1.   3x*-f6a;3  —  5x  —  3  =  0isa  reciprocal  equation  ;  for 

-3-5       6         3 

Ex.  2.  3x*-|-5x3-}-2ic  —  Ssc- 3  =  0is  not  a  reciprocal  equa- 
tion ;  for 

A^A  =  ^  =  ^  =  _i,but2  =  +  i. 

-3-5       5         3  2 

This  equation  becomes  reciprocal  when  —6x  and  —  3  are 
changed  to  +  5  a:  and  +  3. 


*  The  term  reciprocal  equation,  without  the  prefix  self,  usually 
means,  in  mathematical  literature,  what  I  have  here  called  a  self- 
reciprocal  equation. 


TRANSFORMATION.  647 

Ex.  3.   Solve  the  equation 

This  is  a  reciprocal  equation.     It  may  be  written  in  the  form 

or,  again,  in  the  form 

3(.  +  iy_5(x  +  l)+l=0, 

1  3   1 

a  quadratic  equation  in  x  +  -,  whose  roots  are  -,  -.     The  problem 

is  now  reduced  to  the  solution  of  the  two  equations 

^  +  1  =  3  Ul. 

X     2'  X     6 

The  four  roots  of  these  equations  are 

450.  Reciprocal  equations  have  the  following  proper- 
ties, which  are  easily  recognized  as  consequences  of  the 
equations  aja^_^  =  a^^^ja^  =  c.  The  formal  proofs  are 
suggested  as  exercises  for  the  student. 

I.   When  c  =  +  1. 

The  coefficients  of  terms  equidistant  from  the  first  and 
the  last  are  equal ;  i.e.  a^  =  a„_^. 

If  the  degree  of  the  equation  be  odd,  it  has  —  1  as  a 
root.  Its  left  member  is  then  divisible  hj  x  -\-l,  and  the 
quotient,  equated  to  zero,  is  a  reciprocal  equation  of  even 
degree. 


648  TRANSFORMATION. 

11.    Whene  =  -1. 

The  coefficients  of  terms  equidistant  from  the  first  and 
the  last  are  numerically  equal  with  contrary  signs;  i.e. 
a,  =  -  a„_,. 

If  the  degree  of  the  equation  be  even,  the  coefficient  of 
the  middle  term  is  zero. 

If  the  degree  of  the  equation  be  odd,  it  has  +  1  as  a 
root,  and  its  left  member  is  divisible  by  x  —  1.  The 
quotient,  equated  to  zero,  is  a  reciprocal  equation  of  even 
degree. 

If  the  degree  of  the  equation  be  even,  it  has  both  + 1 
and  —  1  as  roots,  and  its  left  member  is  divisible  by 
a^  —  1.  The  quotient,  equated  to  zero,  is  a  reciprocal 
equation  of  even  degree. 

EXAMPLES  ex. 

Apply  the  transformations  indicated,  and  solve  completely  the 
following  equations : 

1.  x^  -S  ax^  +  Sa^x  +  a%  =  0. 
Transformation  :  x  =  a(l  +  0). 

2.  (1  +  c^)x^  -  3  ax2  +  3  a2x  -  a3  =  0. 

Transformation  :  x  =  — - — 

1  +  z 

3.  Solve  completely  the  equation 

(1  +  c3)x8  -  3(a  +  &c3)x2  +  3(a2  +  i^cS-^x  -  (a^  +  b^c^)  =  0. 

4.  Apply  the  transformation  x  =  °  "^  ^^  to  the  equation 

x^  -\-px^  +  qx  +  r  =  0, 

and  assign  to  a  and  /3  such  values  as  will  reduce  this  equation  to 

the  form 

Az^  +  B  =  0. 


TEANSFOEMATION.  649 

Discuss  the  following  special  cases  : 

(1)  pq  =  9r;    (2)Spr  =  q^;    (3)3g=p2; 

(4)   (pq  -dry-  4(3  pr  -  q^)  (Sq-  p"^)  =  0. 

Transform  the  following  equations  into  others,  in  each  of  which 
there  shall  appear  no  term  of  the  second  degree  : 

5.  a;3  +  3a:2-9x  + 1  =0.        7.   x4-8x8+24  x2-28x-85=0. 

6.  3x3+15x2+25x+c=0.        8.   x^-Sx^-{-lSx^-lbx-^6=0. 

Transform  into  equations  whose  coefficients  are  all  integral  the 
following : 

9.    x3  +  1  x2  +  i  X  +  f  =  0. 

10.  X*  +  j\X^  +  7^X2  +  3j|^X  -  5^2  =  0. 

Solve  the  following  equations : 

11.  5x3  +  3x2- 3x-6  =  0. 

12.  x*  -  2  x8  -  6  x2  -  2  X  +  1  =  0. 

13.  X*  -  2  ax8  +  2  ax  -  1  =  0. 

14.  x«  -  3  x5  +  2  x*  -  2  x2  +  3  X  -  1  =  0. 
16.   x4  -  2  x8  -  50  x2  +  2  X  +  1  =  0. 

16.  x'  -  4  x6  -  X*  +  x8  +  4  x2  -  1  =  0. 

17.  Prove  that  every  reciprocal  equation  through  division  (if 
necessary)  by  x  +  1,  or  x  —  1,  or  x2  —  1,  may  be  reduced  to  one 
which  is  of  even  degree  and  whose  coefficients  satisfy  the  condition 

Qr  =  an-r- 


650 


CRITERIA   FOR   REAL  ROOTS. 


CHAPTER   XLVII. 


Criteria  for  Real  Roots. 


.  451.  If  f(x)  be  a  rational  integral  function,  and  if  f(a) 
and  f(P)  be  of  opposite  sign,  f(x)  will  vanish  for  so7ne 
value  of  X  between  a  and  jB. 

For,  when  x  passes,  by  continuous  change,  from  a  to  p, 
fix)  varies  continuously  [Art.  442],  and  therefore  passes 
through  all  intermediate  values  between  /(a)  and  f{(S). 
One  of  these  intermediate  values  must  be  zero,  since  /(a) 
and  /(/8)  are  of  opposite  sign. 

The  graph  of  the  function  makes  the  truth  of  this 
proposition  evident  to  the  eye. 
If  MA,  NB  (Fig.  12)  be  the  ordi- 
nates  /(a),  /(^),  the  one  positive, 
the  other  negative,  the  curve,  in 
passing  from  A  to  B,  will  cross 
the  a;-axis  at  some  point  whose 
ordinate  f{x)  is  zero.  There  may 
be  several  such  points  within  the 
interval  MN. 


Fig.  12. 


Ex.   Find  limits  between  which  the  roots  oix^ —  Zx'^ —  x  +  \  =  Q 
lie. 

If/(x)  =  xs  -  3 x2  -  ic  +  1,  then 

/(-I)  =  -2,  /(t))=l,  /(I)  =  -2,  /(3)  =  -2,  /(4)=13. 

Hence  one  root  lies  between  —  1  and  0,  a  second  between  0  and 
1,  a  third  between  3  and  4. 


\ 
CRITERIA   FOR    REAL   ROOTS.  651 

452.  If  /(a)  and  /(/?)  have  opposite  signs,  f{x)  =  0  has 
an  odd  number  of  real  roots  between  a  and  /3,  but  if  f(a) 
and  f(l3)  have  like  signs,  f(x)  —  0  has  either  an  even  num- 
ber of  real  roots  between  a  and  y8,  or  none. 

If  A  and  B  be  the  points  whose  ordinates  are  /(a) 
and  /(^)  respectively,  then  the  graph  of  f{x),  in  passing 
from  A  to  B,  will  cross  the  aj-axis  an  odd  number  of  times 
if  A  and  B  be  on  opposite  sides  of  this  axis,  but  an  even 
number  of  times,  or  not  at  all,  if  A  and  B  be  on  the  same 
side  of  the  a^axis. 

453.  Every  equation  of  odd  degree  has  at  least  one  real 
root,  positive  or  negative,  according  as  the  constant  term  is 
negative  or  positive. 

For,  if  f(x)  be  of  odd  degree,  and  j>„  be  its  constant 
term, 

/(+Qo)=+QO,  /(0)=p„,  /(-Qo)=-oo; 

and  when  p„  is  positive,  f(x)  =  0  has  a  real  root  between 
0  and  —  cc,  but  when  p^  is  negative,  the  root  is  between 
0  and  H-  oo . 

454.  Every  equation  of  even  degree,  whose  constant  term 
is  negative,  has  at  least  two  real  roots,  one  positive,  the 
other  negative. 

For,  if  f(x)  be  of  even  degree,  and  p^  be  its  constant 
term, 

/(+<^)=+^,    /(0)=p„     /(-c»)  =  +  ^; 

and  since  p^  is  negative,  f{x)  =  0  has  one  real  root  be- 
tween 0  and  -|-  oo,  and  a  second  between  0  and  —  oo. 


652  CRITEKIA  FOR  REAL  ROOTS. 

455.  If  an  equation  with  real  coefficients  have  the  com- 
plex root,  a  +  ih,  it  also  has  the  conjugate  complex  root, 
a  -  ip: 

Assume  /(a  4-  ifi)  =  0.     By  Taylor's  theorem, 

/(„  +  h)  =/(a)  +  hf(a)  +  |l/'(a)  +  .... 

The  value  of  /(a  +  ifi)  is  obtained  from  this  formula 
by  writing  ift  for  h]  and  since  the  even  powers  of 
i  =  ±  1,  and  its  odd  powers  =  ±  i,  the  result  is  of  the 
form 

/(a  +  ip)  =  A-{-  iB, 

where  A  and  B  are  real  quantities.     By  hypothesis, 

/(a4-W  =  0; 

.-.  ^  =  0  and  J5  =  0.  [Art.  234] 

But  /(a  —  ip)  can  be  obtained  from  /(a  +  i^)  by  chang- 
ing i  into  —  i ;  therefore, 

f{a  -ib)  =  A-iB  =  0', 

that  is,  a  —  i/S  is  a  root  of  f(x)  =  0.     Thus,  complex  roots 
always  occur  in  conjugate  pairs. 

456.  If  an  equation  with  rational  coefficients  have  the 
irrational  root,  a  +  -^(3,  it  also  has  the  conjugate  irrational 
root,  a  —  -y/jS. 

By  writing  ^/3  in  place  of  i^,  the  proof  of  this  propo- 
sition becomes  identical  with  that  of  the  preceding 
article. 


CRITERIA  FOR   REAL  ROOTS.  663 

457.  An  equation  whose  coefficients  are  integers,  and  the 
coefficient  of  whose  highest  term  is  unity,  cannot  have  a 
rational  fraction  as  a  root. 

For,  if  possible,  suppose  the  equation 

a;"  =  —  p^a;"-^  —  pg^""^  —  *  •  •  —  i>« 

to  have  the  root  a/^,  in  which  a  and  (3  are  prime  to  one 
another.  Then,  substituting  a/p  for  x,  and  multiplying 
by  ^-\ 

^  =  -p,a--'p  -p^^-'p' p„l3--\ 

an  impossible  result,  since  an  irreducible  fraction  cannot 
be  equal  to  an  integer. 

458.  Descartes'  Rule.    When  in  a  sequence  of  terms,  or 

expressions,  their  signs,  H ,  are  ranged  in  their  proper 

order,  they  are  said  to  present  a  series  of  permanences  and 
variations;    a  permanence  when  two  like   signs   are   in 

juxtaposition,  as  +  +,  or ,  a  variation  when  two 

unlike  signs  are  in  juxtaposition,  as  +  — ,  or  —  -|-. 
Thus  the  sequence  +  —  —  4-  -f  —  has  two  perma- 
nences and  three  variations.  The  following  theorem, 
known  as  Descartes'  Kule  of  Signs,  counts  the  number  of 
variations  presented  by  the  terms  of  an  equation  in  order 
to  give  information  concerning  the  number  of  its  real 
roots.     It  refers  only  to  equations  with  real  coefficients. 

An  equation  f(x)=0  cannot  have  more  positive  real  roots 
than  f(x)  has  variations  of  sign,  nor  more  negative  reaX 
roots  thanf{—  x)  has  variations  of  sign. 


654  CRITERIA    FOR   REAL   ROOTS. 

Consider  any  multinomial  having  both  permanences 
and  variations,  as  +  ^  —  —  —  +  +  —  -|-  —  _  -f. 
We  introduce  a  positive  root  by  multiplying  the  multi- 
nomial by  a  binomial  whose  signs  are  +  — .  The  signs 
involved  in  this  multiplication  range  themselves  thus : 


+ 

+ 

— 

— 

— 

+ 

+   - 

-   +   - 

— 

4- 

+ 

— 

H- 

+ 

— 

— 

— 

+ 

-f   - 

-   4-   - 

— 

+ 

— 

— 

+ 

+ 

+ 

—   - 

-    4-   - 

+ 

+   - 

4-±-T=F4-±-H =FH 

The  double  signs  indicate  ambiguities,  uncertainties. 
Here  are  the  signs  of  the  multiplicand  and  product 
arranged  in  groups,  separated  by  asterisks,  to  call  atten- 
tion to  the  variations  in  the  multiplicand : 

4-±       -TT       -h    ±       -       +      -    T       +   - 
Note  the  following  particularities : 

(1)  In  the  product,  every  ambiguity,  or  group  of 
ambiguities,  has  adjacent  to  it,  -|-  on  one  side,  —  on  the 
other. 

(2)  Therefore,  every  sequence  of  signs  enclosing  such 
an  ambiguity  (or  group),  as-f  ±  — ,  or—  q=  q=  -|-, 
presents  at  least  one  variation,  in  whatever  way  the 
ambiguities  are  interpreted. 

(3)  To  every  variation  in  the  multiplicand  corre- 
sponds, in  the  product,  either  a  variation,  or  an 
ambiguity. 

(4)  A  new  variation  (not  represented  in  the  multi- 
plicand) is  introduced  at  the  end  of  the  product. 


CRITERIA  FOR  REAL   ROOTS.  655 

It  follows  at  once  that  the  variations  in  the  product 
exceed  those  in  the  multiplicand  by  at  least  one.* 

But  we  may  suppose  that  all  the  positive  real  roots 
of  an  equation  are  introduced  in  the  manner  above  in- 
dicated, namely,  through  multiplication  by  a  binomial 
whose  signs  are  -\ — .  Hence  an  equation  cannot  have 
more  positive  real  roots  than  its  terms  have  variations 
of  sign. 

Since  the  negative  roots  of  f(x)  =  0  are  the  positive 
roots  of  f(—x)  =  0,  the  second  part  of  Descartes'  rule  is 
merely  the  first  part  restated  for  the  equation  J{—x)=0. 
Thus  the  above  demonstration  suffices  for  both  parts  of 
the  rule. 

Ex.  1.  The  equation  a^  +  5a;2-f3x  —  7=0  has  certainly  one 
positive  and  one  negative  real  root,  by  Art.  454.-    For  positive 

values  of  x  its  signs  are  +  +  H — ,  and  for  negative  values  +  H . 

Hence,  by  Descartes'  rule,  it  has  only  one  positive,  and  only  one 
negative  real  root. 

Ex.  2.  The  equation  x^  -^  qx  +  r  =  0^  if  q,  r  be  both  positive, 
has  only  one  real  root.  For  if  x  be  positive,  the  signs  are  +  -f  +> 
and  if  x  be  negative,  they  are 1-.    The  real  root  is  negative. 

459.  The  following  propositions  are  consequences  of 
Descartes'  rule  of  signs.  The  proofs  of  them,  here 
omitted,  are  suggested  as  exercises  for  the  student. 

1.  Whe7i  all  the  roots  of  an  equation  f(x)  =  0  are  real, 
the  number  of  its  positive  roots,  and  of  its  negative  roots, 
is  exactly  equal  to  the  number  of  variations  presented  by 
f(x)  and  by  f{—  x)  respectively. 

*  Possibly  by  more  than  one,  always  by  an  odd  number. 


656  CRITERIA  FOR  REAL  ROOTS. 

J 

2.  If  the  terms  of  f(x)  are  all  positive,  the  equation 
f(x)  =  0  has  no  positive  real  root. 

3.  If  all  the  coefficients  of  the  even  poiuers*  of  x  in 
f(x)  are  positive  (or  all  negative)  and  all  the  coefficients  of 
the  odd  powers  of  x  are  negative  (or  all  positive),  the  equa- 
tion f(x)  =  0  has  no  negative  real  root. 

4.  If  f(x)  contain  no  odd  powers  of  x  and  the  coeffir 
cients  are  all  positive^  (or  all  negative),  the  equation 
f(x)  =  0  has  no  real  root. 

5.  If  f(x)  contain  no  even  powers  *  of  x  and  the  coeffi- 
cients are  all  positive  (or  all  7iegative),  f(x)  =  0  has  the  real 
root  x  =  0,  but  no  other. 

EXAMPLES  OXI. 

One  root  of  each  of  the  following  equations  is  indicated ;  find 
all  the  others : 

1.  60a;8-55a;2  +  34a;  +  6=:0;  f(l-V^). 

2.  «*  +  4  a;3  -  5  x2  +  6  X  -  1  =  0  ;  i  (1  -  V^=^). 

3.  4x*-9x2+'3x  +  l=0;  ^(l+\/2). 

4.  a;5-4ic4  +  9x3-x2  +  4x-9  =  0;2+  V^^. 

Assign  a  superior  limit  to  the  number  of  real  roots  which  each 
of  the  following  equations  may  have,  under  the  conditions  enu- 
merated : 

6.    x^  +  qx  +  r  =  0;q>0. 

6.  x^  +  qx  +  r  =  0;  g  <  0. 

7.  £C*  +  5x2  +  rx  +  s  =  0  ;  Q'  >  0. 

8.  x^  ±  (ax^  +  6)  =  0  ;  a  >  0,  6  >  0. 

9.  x^  ±  (ax2  _  6)  =  0  ;  «  >  0,  6  >  0. 

10.   x"^  +  ax*  +  6x3  +  c  =  0  ;  a,  b,  c  unrestricted. 

*  It  is  understood  that  the  constant  term  is  the  coefficient  of  the 
even  power  x^. 


CRITERIA   FOR   REAL  ROOTS.  657 

Determine,  under  the  conditions  indicated,  the  exact  number 
of  the  positive  and  negative  real  roots  of  each  of  the  following 
equations : 

11.  x/^  +  (qx^  +  rx)-s  =  0;  g  >  0,  r  >  0,  s  >  0. 

12.  x*  ±  (qx^  -\-rx)-s  =  0;  g  >  0,  r  >  0,  s  >  0. 

13.  x^  +  (a  -  2)x2  -  2  (a  -  2) X  +  2  a  =  0  ;  a>2. 

14.  x2«+i  +  ax2«-i  +  6x2  +  1  =  0  ;  a>0,  b>0. 

15.  ic2«  +  ax*  +  6x3  -  1  =  0  ;  qj  >  o. 

For  each  of  the  real  roots  of  the  following  equations  find  con- 
secutive integers  between  which  the  root  lies  : 

16.  3  x3  -  8  x2  +  X  +  2  =  0.  18.  x5  +  5  x2  -  3  =  0. 

17.  x4  + 3x2 -60  a: +  34  =  0.  19.   x5  +  5x2  +  3  =  0. 

20.   x7  +  7  x4  +  3  x3  -  1  =  0. 


2n 


658  Sturm's  theorem. 


CHAPTER   XLVIII. 
Sturm's  Theorem. 

460.  The  process*- through  which  an  accurate  account 
of  the  character  and  position  of  the  roots  of  an  equation 
having  real  coefficients  may  h6  rendered  was  discovered 
by  the  French  mathematician  Charles  Sturm,  who  made 
known  his  discovery  to  the  Academy  of  Sciences  at  Paris 
in  1829.*  The  results  of  this  investigation  are  embodied 
in  what  is  known  as  Sturm^s  Theorem. 

461.  Let  the  rational  integral  function  of  x,  of  the  n^^ 
degree,  with  real  coefficients,  be  denoted  by  X,  its  first 
derivative  by  Xi.  The  equation  X  =  0  shall  have  no 
multiple  root,  and  consequently  X  and  X^  shall  have  no 
common  factor  [Art.  437].  Divide  X  by  Xj,  representing 
the  quotient  by  Qi,  the  remainder  by  —  Xg;  then  divide 
Xi  by  Xg,  representing  the  quotient  by  Q2,  the  remainder 
by  —  X3.  Eepeat  this  process  of  dividing  the  last  divisor 
by  the  last  remainder  with  its  signs  changed,  until  a  numeri- 
cal remainder  —  X^  (not  containing  x)  is  obtained.  These 
successive  divisions  produce  j)—l  quotients,  Qi,  Q2,'"  Qp-i, 
and  j9  — 1  modified  remainders,  X2,  Xg,  •••  X^,  viz.  the  true 

*  Published  in  Memoires  presentes  .  .  .  par  des  Savants  Etran^ 
gers,  Paris,  1836. 


stukm's  theorem.  659 

remainders  with  all  their  signs  changed.     The  series  of 
expressions, 

(i).  X,  Xi,  X2,  Xg,  •••  Xp, 

are  called  Sturm's  Functions.    Since  in  the  division  process 
dividend  =  divisor  x  quotient  +  remainder, 

these  functions  and  the  quotients  Qi,  •  •  •  Q^_i  satisfy  the 
identical  relations : 

X    =XM-x^ 

Xi    =  X2V2  —  x^j 
(ii.)  X,    =X,Q,-X^ 


Xp_2  =  Xp_iQp_i  —  Xp. 

The  essential  properties  of  Sturm's  functions  which 
render  them  valuable  for  our  purpose  are  the  following : 

1.  They  are  all  (except  the  last)  continuous  functions 
of  a;  [Art.  442]. 

2.  The  last,  Xp,  being  independent  of  x,  preserves  its 
algebraic  sign  unchanged  whatever  variations  x  may 
undergo. 

3.  Two  consecutive  functions  cannot  become  zero  for 
the  same  value  of  x ;  for  if  two  consecutive  functions  be 
zero,  all  are  zero,  and  X,  Xi  have  a  common  factor. 

4.  If,  for  a  given  value  of  x,  any  one  function,  as  X^, 
vanish,  the  two  adjacent  functions,  X^.i,  X^+i  (on  either 
side  of  X^),   have  opposite  signs ;   for  if  X^  =  0,  then 

5.  Whenever  x,  while  increasing,  passes  through  a 
root  of  X  =  0,  the  ratio  X/X^  changes  from  a  negative 
to  a  positive  value. 


660 

A  formal  proof  of  the  fifth  proposition  is  required: 
Since  X  and  Xi  cannot  both  vanish  for  the  same  value 
of  X,  zero  can  be  neither  a  maximum  nor  a  minimum 
value  of  X  [Art.  438],  and  therefore,  in  passing  through 
zero,  either  X  is  first  negative,  then  zero,  then  positive, 
or  it  is  first  positive,  then  zero,  then  negative.  In  the 
former  case,  X  is  an  increasing  function,  and  Xj  is  posi- 
tive ;  in  the  latter  case,  X  is  a  decreasing  function,  and 
Xi  is  negative  [Art.  438];  and  therefore  in  both  cases 
the  ratio  X/Xi  passes  from  a  negative  value  through 
zero  to  a  positive  value. 

462.  Sturm's  functions  are  applied  to  the  determinar 
tion  of  the  number  of  the  real  roots  of  an  equation, 
X  =  0,  that  lie  between  any  two  numbers,  a  and  ^ 
LP  >  a],  in  the  following  manner : 

In  each  of  Sturm^s  functionSy 

X,  Xi,  X2,  •••  Xp_i,  Xp, 

substitute  a  for  x,  producing  p  -f  1,  positive  and  negative, 
numbers.  Count  the  variations  in  sign  that  occur  in  pass- 
ing from  the  first  to  the  second,  to  the  third,  .  .  .  to  the  last, 
of  these  numbers,  and  denote  the  number  of  variations  by 
v^.  Substitute  similarly  p  for  x,  count  again  the  variations 
in  sign,  and  denote  their  number  by  v^.     The  difference 

is  the  exact  number  of  the  real  roots  of  X=  0,  that  lie 
between  a  and  ft.     This  is  Sturm's  Theorem. 

To  prove  this  theorem,  let  x  be  supposed  to  pass  by 
continuous  increase  from  a  to  p.  Since  X,  Xi,  X2,  •••, 
Xp_i,  are  all  continuous  functions  of  x,  none  of  them  can 


x„ 

^r+l 

.     In  fact, 

thef( 

Before  the  change. 

^.-1 

Xr 

Xr^i 

(1) 

— 

+ 

+ 

(2) 

— 

— 

+ 

(3) 

+ 

+ 

— 

(4) 

+ 

— 

— 

Sturm's  theorem.  661 

change  sign  without  passing  through  zero,  and  the  sign 
of  Xp  is  invariable.  But  when  any  one  except  the  first 
becomes  zero,  say  X^  =  0,  the  two  adjacent  functions, 
X^_i,  X^^i,  have  opposite  signs,  and  a  change  of  the  sign 
of  X^  from  -f  to  — ,  or  from  —  to  +,  cannot  affect  the 
number  of  variations  presented  by  the  three  terms  X^_^ 

possible  alternatives  are: 

After  the  change. 

Xr-i  X^  X^+i 

-  -  + 

-  -h  + 
+  -  - 

+  +  - 

In  every  instance,  one  permanence  and  one  variation. 
Hence  no  variations  are  gained  or  lost  through  any 
changes  of  sign  that  can  occur  in  the  functions  Xi,  Xg, 
-Xs)  "'j  Xp. 

Let  X  now  pass  through  a  root  of   X=0.     At  this 

instant  X  changes  sign,  and  the  ratio  X/X^  passes  from 

a  negative  to  a  positive  value.    Hence  only  four  different 

groupings  of  the  signs  of  X,  X^  X2,  before  and  after  x 

passes  the  root  of  X  =  0,  are  possible,  namely  : 

Variations 
Before.  After,  lost. 

(1)  +         -         +  -         -         +  1 

(2)  +       -       -  _       _       _  1 

(3)  -       +       4-  +       +       4-  1 

(4)  -       +       -  +       +       _  1 

In  every  case  one,  and  only  one,  variation  is  lost ;  and, 
since  the  vanishing  and  consequent  changing  of  sign  of 


662 


STURM  S   THEOREM. 


the  ratio  X/X^  presents  the  only  readjustment  of  signs 
that  can  cause  the  disappearance  of  variatigjis  in  sign, 
no  such  disappearance  will  take  place  except  when  x 
passes  a  root  of  X  =  0.  Hence  the  number  of  real  roots 
of  X  =  0  between  a  and  y8  is  exactly  v^  —  v^.     q.e.d. 

Hence,  the  total  number  of  real  roots  of  an  equation^ 
X=0,  is 

463.  If  fractional  coefficients  make  their  appearance 
in  Sturm's  process,  they  may  be  got  rid  of  by  introducing 
or  removing  suitable  positive  factors,  as  is  done  in  finding 
highest  common  factor  [Art.  135].  When  the  process  is 
thus  adulterated,  Sturm's  functions  are  correspondingly 
modified,  but  without  losing  any  of  their  fundamental 
characteristics  described  in  Art.  461. 


Ex.  1.   Determine  the  number  of  the  real  roots  of  the  equation 

a;3  -  5  jc  +  5  =  0. 

We  have  X=  x^  —  5  x  +  5,  Xi  =  3x2  —  5.  The  division  pro- 
cess, for  the  determination  of  X2,  X3,  etc.,  may  be  exhibited  as 
follows : 


X  = 


x^  -    5x+    5 
3 

3x3-15x  +  15 
3x8-    5x 

3x2-    5 
2 

lOx-15 
X2=    2x-    3 

6x2-  10 
6  x2  -  9  X 

9x  -  10 
9x  -  ^ 

Xi 


3x+f  =  $2 


'-■^+   10  =  -|  =  X8 


Sturm's  theorem. 


Thus  Sturm's  functions  (modified),  four  in  number,  are 

x3_5x  +  5,   3x2-5,   2x-3,    -  |. 
The  following  table  of  variations  shows  that  the  equation  X=  0 
^  only  one  real  root,  a  negative  one,  between  —  2  and  —  3  : 
Value  of  %.  Sequence  of  Signs.  Variations. 

-00  _      +      _      _  2  * 

+  00  +       +       +       _  1 

-3  _     +     _     _  2 

-2  +     +     _     _  1 

The  following  is  a  graph  of  the  function  X: 


y  =  aj8  —  5x  +  5 


X 

y 

-4 

-39 

-3 

-    7 

-2 

7 

-1 

9 

0 

5 

1 

1 

2 

3 

3 

17 

Fig.  13. 


Until  some  test  is  applied,  it  is  doubtful  whether  the  curve 
crosses  the  horizontal  axis  between  the  abscissae  1  and  2.  The 
test  by  Sturm's  functions  shows  that  it  does  not.  By  Horner's 
process,  the  real  root  is  found  to  be  —  2.627+. 

By  the  criterion  for  maxima  and  minima,  [Art.  438],  the  co- 
ordinates of  the  elbow-points,  J.,  i?,  in  the  graph,  are  —  f, 
5+10v'i^5,  and  f,  5  —  lO^j^,  and  this  result  also  shows  that 
the  graph  does  not  cross  the  a;-axis  a  second  time. 

464.  In  order  that  all  the  roots  of  an  equation  of  the 
wth  degree  may  be  real  and  distinct,  it  is  necessary  and 


664  Sturm's  theorem. 

sufficient  that  the  series  of  Sturm's  functions  shall  lose 
exactly  n  variations  when  x  passes  from  —  oo  to  -f  oo. 
There  must  therefore  be  just  n  +  1  Sturm's  functions, 

whose  degrees  are  respectively  n,  n  —  1,  "•  2,  1,  0,  alter- 
nately even  and  odd;  and  this  series  must  present,  for 
ic  =  —  00  ^  variations,  for  ic  =  -h  oo  n  permanences,  since 

we  must  have 

"W-oo  —  y+00  =  n. 

But  this  requires  that  the  coefficient  of  the  first  term 
(term  of  highest  degree)  of  every  one  of  the  functions 
X,  •••  X„  shall  be  positive.     Hence: 

In  order  that  an  equation  of  the  n^^  degree  may  have 
all  its  roots  real  and  distinct,  it  is  necessary  and  sufficient 
that  its  Sturm^s  functions  shall  he  n-\-l  in  number,  and 
shall  have  positive  numbers  only  as  coefficients  of  their 
terms  of  highest  degree. 

Ex.  1.  The  equation  x^  —  6x  +  3  =  0  has  all  its  roots  real; 
between  what  integers  do  they  lie  ? 

Sturm's  functions  (modified)  are  : 

a;3_5a;  +  3,   3a;2  +  6,   lOx-9,   \5^; 
four  in  number,  coefficients  of  terms  of  highest  degree  all  positive. 
The  sequences  of  sign  for  aj  =  —  3,  —  2,  0,  1,  2  are  : 


Value  of  X. 

Sequence  of  Signs. 

Variations. 

-3 

-     +     -     + 

3 

-2 

+     +     -     + 

2 

0 

+     -     -     + 

2 

1 

-     -     +     + 

1 

2 

+     +.     +     + 

0 

Thus  one  root 

lies  between  -  2  and  —  3, 

another  between  0 

and  1,  a  third  between  1  and  2. 

Sturm's  theorem.  665 

465.  If  among  Sturm's  functions  an  intermediate  one, 
as  X^,  is  found,  which  cannot  change  its  sign  when  x 
ranges  through  the  values  to  be  assigned  to  it,  the  func- 
tions which  follow  it  may  be  disregarded ;  the  computa- 
tion may  stop  with  X^.  The  functions  X,  Xi,  Xg,."-  X^ 
then  determine  the  number  and  position  of  the  real  roots  ; 
for  they  have  all  the  fundamental  properties  which  suf- 
fice to  prove  Sturm's  theorem. 

Ex.  1.  Thus,  for  the  equation  x^  +  Bx^-\-k  =  0  the  only  func- 
tions that  need  be  considered  are 

X=(i(^  +  6ci^  +  k,   Xi  =  6(x^+  3 x2), 

for  X*  4-  3  x2  cannot  change  its  sign  for  real  values  of  x.    The  appli- 
cation of  the  test  gives 

V-oo  —  V+oe  =1  —  0  =  1. 

The  equation  has  only  one  real  root. 

EXAMPLES   CXII. 

Determine,  by  Sturm's  theorem,  the  number  and  position  of 
the  real  roots  of  the  following  equations  : 

1.  x8-3x2  +  3a:-ll  =  0.  6.   x'^  +  ix^  +  ix^  +  48  =  0. 

2.  x^-Zx^-3x  +  ll=0.  6.   x5-5x-l  =  0. 

3.  .x3  -  2  x2  -}-  X  -  1  =  0.  7.  x*  +  3  x*  -f  3  x2  -I-  a2  =  0. 

4.  2x8-7ax2+10a2x-6a3=0.      8.   x"^  -  7  x^  4- 1  =  0. 
9.    Show  that,  if  a  and  h  be  positive  real  numbers, 

x^  -1-  5  ax^  +  5  6x  -f  c  =  0 
has  one,  and  only  one,  real  root. 

10.    Show  that,  if  a,  m,  and  n  be  positive  real  numbers, 
X2m+1  _|_  ax2'»+i  4-5=0 

has  one,  and  only  one,  real  root. 


666 


MISCELLANEOUS   EXAMPLES   VIII. 


11.  Show  that  two  of  the  roots  of  x'^— 4iX  +  Sa  =  0  are  neces- 
sarily imaginary,  and  that  the  other  two  are  real  if  a<l,  or 
a  =  1,  imaginary  if  a  >  1. 

12.  Show  that 

x4  _  2  x3  -  (2  (z  -  3)  a;2  +  2  (a  -  l)x  +  (a  -  1)  (a  -  2)  =  0 
has  all  its  roots  real  if  a  >  1,  or  a  =  1,  all  its  roots  imaginary  if 
a<l. 

13.  Show  that 

x*  -  2  (a  +  l)x3  +  (a2  +  4  rt  +  1)  x2  -  2  (a3  +  1) a;  +  a2  =  0 
always  has  two  real  and  two  imaginary  roots  except  when  a^  =  I. 

14.  Obtain  Sturm's  functions  for  the  cubic  x^  +  qx  -\-  r  =  0,  and 
show  that  the  condition,  necessary  and  sufficient,  that  the  three 
roots  of  this  equation  shall  be  real  is  4:q^  -{■  27  r^<iO. 

15.  Making  use  of  the  results  of  Example  14  and  of  the  trans- 
formation of  Art.  408,  obtain  Sturm's  functions  for  the  general 
cubic  x^  +  px^^  -{-  qx  -}-  r  =  0. 

16.  Obtain  Sturm's  functions  for  the  equation  x""  —  x-{-  1  =0 
and  show  that  it  has  only  one  real  root  if  n  is  odd,  none  if  n  is  even. 

MISCELLANEOUS   EXAMPLES  VIII. 


1.   Verify  that 

1     a    a2 

= 

1    a 

be 

- 

1     6     62 

1    b    ca 

1    c     c2 

1    c    ab 

. 

2.   Prove  that 

-  6c       6c  +  62     6c  +  c2 

, 

ca  +  a2       —  ca      ca  +  c2 

a6  +  a2    a6  +  62      -  aU 

=  2 

(b  +  C)2              C2                      62 

c2          (c  +  ay         a2 

=  2  (6c 

62 

( 

a  - 

hby 

3.    Solve  the  equation 

x  —  a  x  —  b  X  —  c 

x  —  b  X  —  c  X  —  a 

X  —  c  x  —  a  x—b 


=  0. 


MISCELLANEOUS   EXAMPLES  Vm.  667 


4.   Verify  tliat 

a-^      a^      a^ 
a\bi    aihi    a^hz 
6i2       bi^       632 

=  (a2&3- 

-  azbi)  (asfti  -  ai&s)  (ai&2 

a2&i). 

Find  the  rational  linear  factors  of  the  following  expressions : 

5.  x\y  +  2!)  +  y\z  +  x)  +  z\x  +  ?/)-(«/  +  0)  (2  +  x)  (x  +  t/). 

6.  X3  +  y3  +  ^3  _|_  2  x?/0  -  X2(j/  +  2;)  -  2/2(5;  +  x)  -  z\x  +  y). 

7.  2(1/202  +  ;22aj2  ^.  ^.2^2)  _  a;4  _  y4  _  ;j4. 

8.  Show  that  each  of  the  linear  factors  of 

X2(X2  -  4  yt  -  22)  _  y2(y2  _  4  X0  -  «2) 

+  22(^2  _  4  y^  _  a;2)  _  ^2(^2  _  4  2;x  -  y2) 

is  of  the  form  % -^  y\ -\- z>? -\- 1\^,  where  X*  -  1  =  0. 

9.  Show  that  if  ax^  +  2hxy  +  by^ -^2  gx -^  2fy  +  c  he  factor- 
able and  one  factor  is  x  +  Xy  +  /*,  the  other  is 

ax-\-(2h  —  Xa)?/  +  2  g  —  fxa. 

10.  Prove  that  if  x*  +  px^  +  qx^  -\- rx -\-  s  =  0  have  three  equal 
roots,  q"^  -  9  pr  -{- 12  s  z=  0. 

11.  Given,  that  a,  /3,  7  are  the  roots  of  x^  +px2  +  gx  +  r  =.'  y, 
write  the  equation  whose  roots  are  : 

(1)  a2,  )82,  72;    (2)   l/a2,  I//32,  I/72. 

12.  Given,  that  a,  /S,  7  are  the  roots  of  x^  4-l>x2  +  gx  +  r  =  0, 
find  the  roots  in  terms  of  o,  /3,  7,  of 

r2x8  +  (2  qr  -p^)x^  +  (q^  -  2pr)x  +  r  =  0. 

13.  The  roots  of  x^  +px^  +  gx  +  r  =  0  being  a,  j3,  7,  express  the 
cyclo-symmetric  function  a2/3  +  /32«y  _|.  ^2^  Jq  terms  of  p,  g,  and  r. 

14.  Solve  the  equations  z^  =  1  and  2;2o  z=  —  1. 

15.  Show  that  ax"  +  &  =  0  may  be  expressed  in  the  form  of  a 
reciprocal  equation. 

16.  Given  x2  —  2;x  +  1  =  0  and  f(z)  =  x>  +  1,  prove  that  all  the 
roots  of  f{z)  =  0  are  real  and  are  situated  between  +  2  and  —  2. 


668  MISCELLANEOUS   EXAMPLES  YUL, 

17.  Construct  the  graph  of  (x^  —  l)/x^. 

18.  Find  the  maximum  and  minimum  values  of 

2  x3  +  3  x2  -  36  X  -  10. 

19.  Determine  the  shape  of  the  rectangle  of  maximum  area  that 
caii  be  inscribed  in  a  circle  of  giren  radius. 

20.  Determine  the  shape  of  the  rectangle  of  maximum  area 
that  can  be  inscribed  in  a  given  square. 

21.  Determine  the  shape  of  a  triangle  of  given  area  and  mini- 
mum perimeter.  • 

22.  Solve  completely  the  equation 

(1  -  c4)x4  -iax^  +  Q  d?-x^  -  4  a^x  +  a*  =  0. 

23.  Solve  completely  the  equation 

(1  -  c*)x4  -  4  (a  -  5c*)x3  ^^{a^-  hH'^)x^  +  4  (a^  -  h^cS)x 

4-(a*-&4c4)=o. 

24.  Find  the  relation  between  the  coefficients  of 

X*  +  px8  +  gx2  +  rx  +  s  =  0, 
y/Ji^ich  will  make  it  possible  to  reduce  this  equation  to  the  form 


25.   Solve  the  simultaneous  equations 


—  a     a  —  b      a  —  c 


y  —  a     y  —  b     y  —  c 
26.   Prove  that  all  the  roots  of  the  equation 


X  —  a     X  —  /3     X  —  y  X—  V 

are  real. 


MISCELLANEOUS  EXAMPLES  VIH.  669 

27.  Show  that  the  three  roots  of 

—  a^(x  —  a)  —  h'^{x  —  jS)  —  c'^(x  —  7)  —  2  ahc  =  0    are  real. 

28.  Prove  that 

/W^^_.  _1_  .^_  ,        .  ^_    . 
f(x)      x-a'^x-  ^^x-y'^    '^  x-v 

where  a,  /3,  7,  ...  v,  are  the  roots  of  f{x)  =  0. 

29.  Given,   thsU  f(x)  =  ^^~^^,  and  that /(«)  (x)  =  Xn  =  n*!^ 
derivative  of /(x),  verify  that 

(l-x2)/(x)+2wx/(x)=0, 

(1  -  x^)Xn"  -  2  xX„'  +  n(n  +  1)X„  =  0, 

where  X„",  X„'  =  first  and  second  derivatives,  respectively,  of  X„. 

30.  Apply   the    transformation    x  =  "^  "^  ^    to    ax^  +  2bx  +  c 

yz+  S 

and,  denoting  the  transformed  function  by  a'z^  +  2b'z  +  c',  prove 
that 

a'cf  -  6'2  =(a5  -  pyy(ac  -  62). 

31.  Apply  the  transformation  x  =^£±E  to  ax^+Shx^+Zcx-td; 

yz+8 
and,  denoting  the  transformed  function  by  a'z^  +  3  b'z"^  +  Sc'z  -^  d', 
prove  that 

a'2d'2  +  4  a'c'3  +  4  d'6'8  -  3  6'2c'2  -  6  a'b'c'd' 

=  (ad  -  pyy(a'^d^  +  4  ac3  +  4  d&3  -  3  62c2  _  6  abed). 


670 


TABLE   OF   LOGARITHMS. 


N. 

0    12   3   4 

6   6   7    8   9 

d. 

10 

12 
13 

0000  0043  00S6  0128  0170 

0212  0253  0294  0334  0374 

41 

0414  0453  0492  0531  0569 
0792  0828  0864  0899  0934 
1139  1 173  1206  1239  1271 

0607  0645  0682  0719  0755 
0969  1004  1038  1072  1 106 
1303  1335  ^3^7   1399  1430 

38 

35 
32 

'14 

1461  1492  1523  1553  1584 
1761  1790  1818  1847  1875 
2041  2068  2095  2122  2148 

1614  1644  1673  1703  1732 
1903  1931  1959  1987  2014 
217;^  2201  2227  2253  2279 

^0 
28 

26 

17 
18 

19 

20 

21 

22 
23 

2304  2330  2355  2380  2405 
2553  2577  2601  2625  2648 
2788  2810  2833  2856  2878 

2430  2455  2480  2504  2529 
2672  2695  2718  2742  2765 
2900  2923  2945  2967  2989 

25 
24 
22 

3010  3032  3054  3075  3096 

3118  3139  3160  3181  3201 

21 

3222  3243  3263  3284  3304 
3424  3444  3464  3483.  3502 
3617  3636  3655  3674  3692 

3324  3345  3365  3385  3404 
3522  3541  3560  3579  3598 
37"  3729  3747  3766  3784 

20 

19 
18 

24 

3802  3820  3838  3856  3874 

3979  3997  4014  4031  4048 
4150  4166  4183  4200  4216 

3892  3909  3927  3945  3962 
4065  4082  4099  41 16  4133 
4232  4249  4265  4281  4298 

18 

17 
16 

27 
28 
29 

30 

31 
32 
33 

4314  4330  4346  4362  4378 
4472  4487  4502  4518  4533 
4624  4639  4654  4669  4683 

4393  4409  4425  4440  4456 
4548  4564  4579  4594  4609 
4698  4713  4728  4742  4757 

16 
15 

15 

4771  4786  4800  4814  4829 

4843  4857  4871  4886  4900 

14 

14 
13 
13 

4914  4928  4942  4955  4969 
5051  5065  5079  5092  5105 
5185  5198  5211  5224  5237 

4983  4997  50"  5024  5038 
5119  5132  5145  5159  5172 
5250  5263  5276  5289  5302 

34 
35 
36 

5313  5328  5340  5353  5366 
5441  5453  5465  5478  5490 
5563  5575  5587  5599  561 i 

5378  5391  5403  5416  5428 
5502  5514  5527  5S39  5551 
5623  5631  5647  5658  5670 

13 
12 
12 

37 
38 
39 
40 

5682  5694  5705  5717  5729 
5798  5809  5821  5832  5843 
591 1  5922  5933  5944  5955 

5740  5752  5763  5775  5786 
5855  5866  5877  5888  5899 
5966  5977  5988  5999  6010 

12 
II 
II 

6021  6031  6042  6053  6064 

6075  ^85  6096  6107  61 1 7 

" 

N. 

0   12   3   4 

5   6   7   8   9 

d. 

TABLE   OF   LOGARITHMS. 


671 


N. 

0    12    3    4 

6    6   7    8   9 

d. 

40 

41 
42 

43 

44 
45 
46 

47 
48 

49 
50 

51 
52 
53 

54 

55 
56 

11 

59 
60 

61 
62 
63 

64 

^1 

69 
70 

6021  6031  6042  6053  6064 

6075  6085  6096  6107  61 1 7 

10 
10 
10 

10 

10 

9 

9 
9 
9 

6128  6138  6149  6160  6170 
6232  6243  6253  6263  6274 
6335  6345  6355  6365  6375 

6435  6444  6454  6464  6474 
6532  6542  6551  6561  6571 
6628  6637  6646  6656  6665 

6721  6730  6739  6749  6758 
6812  6821  6830  6839  6848 
6902  691 1  6920  6928  6937 

6180  6191  6201  6212  6222 
6284  6294  6304  6314^6325 
6385  6395  6405  6415  6425 

6484  6493  6503  6513  6522 
6580  6590  6599  6609  6618 
6675  6684  6693  6702  6712 

6767  6776  6785  6794  6803 
6857  6866  6875  6884  6893 
6946  6955  6964  6972  6981 

6990  6998  7007  7016  7024 

7033  7042  7050  7059  7067 

9 

8 
8 
8 

8 
8 
8 

8 

7076  7084^023.7101  7110 
7160  7168  7177  7185  7193 
7243  7251  7259  7267  7275 

7324  7332  7340  7348  7356 
7404  7412  7419  7427  7435 
7482  7490  7497  7505  7513 

7559  7566  7574  75^2  75^9 
7634  7642  7649  7657  7664 
7709  7716  7723  7731  7738 

7118  7126  7135  7143  7152 
7202  7210  7218  7226  7235 
7284  7292  7300  7308  7316 

7364  7372  7380  7388  7396 
7443  7451  7459  7466  7474 
7520  7528  7536  7543  7551 

7597  7604  7612  7619  7627 
7672  7679  7686  7694  7701 
7745  7752  7760  7767  7774 

7782  7789  7796  7803  7810 

7818  7825  7832  7839  7846 

7853  7860  7868  7875  7882 

7924  7931  7938  7945  7952 
7993  8000  8007  8014  8021 

8062  8069  8075  8082  8089 
8129  8136  8142  8149  8156 
8195  8202  8209  8215  8222 

8261  8267  8274  8280  8287 
8325  8331  8338  8344  8351 
8388  8395  8401  8407  8414 

7889  7896  7903  7910  7917 
7959  7966  7973  798o  7987 
8028  8035  8041  8048  8055 

8096  8102  8109  81 16  8122 
8162  8169  8176  8182  8189 
8228  8235  8241  8248  8254 

8293  8299  8306  8312  8319 
8357  8363  8370  8376  8382 
8420  8426  8432  8439  8445 

7 
7 

6 
6 
6 

8451  8457  8463  8470  8476 

8482  8488  8494  8500  8506 

6 

N. 

0   12   3   4 

5   6   7   8   9 

d. 

672 


TABLE   OF   LOGARITHMS. 


N. 

0    12    3    4 

5    6    7    8    9 

d. 

70 

71 

72 

n 

74 

75 
76 

77 
78 
79 

80 

81 
82 
83 

84 
85 
86 

87 
88 

89 
90 

91 
92 
93 

94 

96 

97 

98 

99 

100 

8451  8457  8463  8470  8476 

8482  8488  8494  8500  8506 

6 

8513  8519  8525  8531  8537 
8573  8579  8585  8591  8597 
8633  8639  8645  8651  8657 

8692  8698  8704  8710  8716 
8751  8756  8762  8768  8774 
8808  8814  8820  8825  8831 

8865  8871  8876  8882  8887 
8921  8927  8932  8938  8943 
8976  8982  8987  8993  8998 

8543  8549  8555  8561  8567 
8603  8609  8615  8621  8627 
8663  8669  8673  8681  8686 

8722  8727  8733  8739  8745 
8779  8785  8791  8797  8802 
8837  8842  8848  8854  8859 

8893  8^99  8904  8910  8915 
8949  8954  8960  8965  8971 
9004  9009  9015  9020  9025 

6 
6 
6 

6 
6 
6 

6 
6 

5 

9031  9036  9042  9047  9053 

9058  9063  9069  9074  9079 

5 

5 
5 
5 

5 
5 
5 

5 
5 
5 

9085  9090  9096  9101  9106 
9138  9143  9149  9154  9159 
9191  9196  9201  9206  9212 

9243  9248  9253  9258  9263 
9294  9299  9304  9309  9315 
9345  9350  9355  936o  9365 

9395  9400  9405  9410  9415 
9443  9450  9455  9460  9465 
9494  9499  9504  9509  9513 

9112  9117  9122  9128  9133 
9163  9170  9175  9180  9186 
9217  9222  9227  9232  9238 

9269  9274  9279  9284  9289 
9320  9325  9330  9333  9340 
9370  9375  9380  9385  9390 

9420  9425  9430  9435  9440 
9469  9474  9479  9484  9489 
9518  9523  9528  9533  9538 

9542  9547  9552  9557  95^2 

9566  9571  9576  9581  9586 

5 

9590  9595  9600  9605  9609 
9638  9643  9647  9652  9657 
9683  9689  9694  9699  9703 

9731  9736  9741  9745  9730 
9777  9782  9786  9791  9795 
9823  9827  9832  9836  9841 

9868  9872  9877  9881  9886 
9912  9917  9921  9926  9930 
9956  9961  9965  9969  9974 

9614  9619  9624  9628  9633 
9661  9666  9671  9675  9680 
9708  9713  9717  9722  9727 

9754  9759  97^3  9768  9773 
9800  9805  9809  9814  9818 
9845  9850  9854  9859  9863 

9890  9894  9899  9903  9908 
9934  9939  9943  9948  9952 
9978  9983  9987  9991  9996 

5 
5 
5 

5 
5 
4 

4 
4 
4 

0000  0004  (X)09  0013  0017 

0022  cx)26  0030  0035  0039 

4 

N. 

0   12   3    4 

5    6   7    8    9 

d. 

\   ■: 


.^S. 


'  J- 


..4-   I 


.^10. 


$\ 


m 


.*.-, 


l-S^'- 


